Free math worksheets

- Assume the circles share a point. - Express the distance from that point to the common center in two ways. - Compare the resulting radius values.

1. Let the circles have common center \(O\) and different radii \(r_1\) and \(r_2\). Assume for contradiction that they share a point \(P\). 2. Since \(P\) lies on the first circle, \(OP=r_1\). 3. Since \(P\) also lies on the second circle, \(OP=r_2\). 4. Therefore, \(r_1=r_2\), contradicting the fact that the radii are different. 5. Hence the two concentric circles have no common point.

A common point would have to be both radii away from the shared center, forcing the radii to be equal. This contradiction shows that the circles do not intersect.

- Start by negating exactly the statement you are trying to prove. - Keep the assumption weaker than any contradiction you hope to derive later.

1. An indirect proof begins by assuming the negation of the conclusion. 2. The negation of “\(\ell\parallel m\)” is “\(\ell\) and \(m\) are not parallel.” 3. For distinct coplanar lines, that means assuming they intersect.

Assume that \(\ell\) and \(m\) are not parallel; for distinct coplanar lines, assume they intersect.

- Negate the quantity phrase “at most one” carefully. - The opposite must include every case in which the conclusion fails.

1. “At most one” allows zero or one obtuse angle. 2. Its negation is that there are more than one. 3. Therefore, assume the triangle has at least two obtuse angles.

Assume that the triangle has at least two obtuse angles.

- Add the lower bounds for the two angles. - Compare that sum with the total available angle measure in a triangle.

1. If both \(\angle A\) and \(\angle B\) exceed \(90^\circ\), their sum already exceeds \(180^\circ\). 2. A triangle's three interior angles must total exactly \(180^\circ\). 3. Therefore, choice B is the fact that contradicts the derived inequalities.

B

- Begin by assuming the conclusion is false. - Focus on the angles each line makes with the same ray at \(A\). - Identify the uniqueness condition that the assumption would violate.

1. Assume for contradiction that \(m\) and \(n\) are distinct lines. 2. Because both lines pass through \(A\), they form an angle at \(A\). 3. Since \(m\perp\ell\) and \(n\perp\ell\), each line forms a \(90^\circ\) angle with the same ray of \(\ell\) at \(A\). 4. From a fixed ray at a point, there is only one ray on a given side that forms a \(90^\circ\) angle with it. Thus the rays of \(m\) and \(n\) on that side must coincide. 5. This contradicts the assumption that \(m\) and \(n\) are distinct. Therefore, \(m=n\).

Assuming \(m\ne n\) produces two different perpendiculars to \(\ell\) through the same point \(A\), contradicting the uniqueness of a \(90^\circ\) ray from a fixed ray at \(A\). Therefore, \(m\) and \(n\) are the same line.

- Use the quadrilateral angle sum exactly as written. - A contradiction must conflict with the statement assumed at the beginning. - After identifying the conflict, negate the contrary assumption to state the conclusion.

1. The angle-sum equation gives \(270^\circ+x^\circ=360^\circ\), so \(x=90\). 2. The proof began by assuming that the fourth angle is not \(90^\circ\), but the angle sum forces it to be \(90^\circ\). 3. That direct contradiction shows the starting assumption is false. 4. Therefore, the fourth angle is a right angle.

a) \(x=90\). b) The assumption says \(x\ne90\), while the angle sum forces \(x=90\). c) The fourth angle must be a right angle.

- Distinguish “at least one” from “at most one.” - A uniqueness proof should assume two different objects both satisfy the defining property. - Points at the same distance from \(A\) on the same ray cannot be distinct.

1. “Exactly one” contains both existence and uniqueness. Assuming there is no midpoint attacks existence, not the uniqueness claim that no two distinct midpoints can exist. 2. For an indirect uniqueness proof, assume there are two distinct midpoints \(M\) and \(N\). 3. Because both are midpoints, \(AM=\frac12AB\) and \(AN=\frac12AB\), so \(AM=AN\). 4. But if \(M\) lies strictly between \(A\) and \(N\), then \(AM<AN\). 5. The statements \(AM=AN\) and \(AM<AN\) contradict each other, so two distinct midpoints cannot exist.

a) “No midpoint” negates existence, not uniqueness. b) Assume there are two distinct midpoints \(M\) and \(N\). c) Both would satisfy \(AM=AN=\frac12AB\), but if \(M\) lies between \(A\) and \(N\), then \(AM<AN\). Contradiction.

- Begin by assuming the prohibited configuration exists. - Use the defining measure condition for each angle type. - Compare the partial angle sum with the total available in a triangle.

1. Assume for contradiction that a triangle has two obtuse angles. 2. Each obtuse angle measures more than \(90^\circ\), so the sum of those two angles is greater than \(180^\circ\). 3. The three interior angles of a triangle have total measure exactly \(180^\circ\). 4. The two obtuse angles alone would exceed the entire allowed angle sum, which is impossible. 5. Therefore, a triangle cannot contain two obtuse angles.

Assuming two obtuse angles makes their sum greater than \(180^\circ\), contradicting the triangle angle-sum theorem. Therefore, a triangle can have at most one obtuse angle.

- Under the assumption, determine the relationship between \(OT\) and line \(m\) in two different ways. - The contradiction should be a statement that no ordinary line can satisfy. - Reject only the temporary assumption, not the tangent-radius theorem.

1. Under the assumption, \(OT\) lies along line \(m\). 2. Tangency also gives \(OT\perp m\). 3. Thus line \(m\) would be perpendicular to itself, which is impossible. 4. Therefore, the assumption that \(m\) passes through \(O\) must be rejected. 5. Hence a tangent line cannot pass through the circle's center.

a) The assumption makes \(m\) perpendicular to itself, an impossibility. b) Reject the assumption that \(m\) passes through \(O\). c) A tangent line to a circle cannot pass through the center.

- Focus on how many parallels to \(\ell\) can pass through one point. - The temporary intersection point \(P\) is what makes the uniqueness theorem applicable. - After finding the contradiction, reject the assumption that \(m\) and \(n\) intersect.

1. The Parallel Postulate gives exactly one line through a point \(P\) parallel to a given line \(\ell\). 2. The assumption produces two distinct lines, \(m\) and \(n\), through \(P\), each parallel to \(\ell\). 3. That contradicts uniqueness of the parallel through \(P\). 4. Therefore, the assumption that \(m\) and \(n\) intersect is false, so \(m\parallel n\).

The contradiction is that two distinct lines through the same point \(P\) would both be parallel to \(\ell\), violating uniqueness of a parallel through a point. Therefore, \(m\parallel n\).

- Correct mathematics and correct proof classification are separate questions. - Look for an explicit temporary assumption that denies the desired conclusion. - The contradiction should be with the definition of a scalene triangle.

1. The mathematical conclusion is correct: congruent angles would imply congruent opposite sides, which a scalene triangle does not have. 2. Itzel’s written argument does not begin by assuming the negation of the conclusion and deriving a contradiction. It is essentially a direct contrapositive-style argument. 3. A genuine contradiction proof would begin: “Assume, for contradiction, that the scalene triangle has two congruent angles.” 4. The angle-side theorem would then force two congruent sides, contradicting the definition of scalene.

a) Yes. b) No. It does not use a temporary contrary assumption and contradiction; it argues directly from the triangle’s side condition. c) Begin: “Assume, for contradiction, that the scalene triangle has two congruent angles.”

- Combine the original rhombus properties with the new rectangle properties established in step 3. - Focus on which defining properties are now all true at once. - The missing statement should directly conflict with the temporary assumption.

1. Step 3 establishes that the given rhombus is also a rectangle. 2. A quadrilateral that is both a rhombus and a rectangle has four congruent sides and four right angles, so it is a square. 3. Therefore, step 4 is: “The rhombus is a square.” 4. That statement directly contradicts the temporary assumption that it is not a square. 5. Hence the original rhombus with congruent diagonals must be a square.

a) Step 4: “The rhombus is a square.” b) It is both a rhombus and a rectangle, so it has the defining properties of a square. c) A rhombus with congruent diagonals is a square.

- Translate the negation of “four congruent sides” carefully. - Use both pairs of opposite sides of the parallelogram. - Compare every side to the given congruent consecutive pair.

1. If the parallelogram were not a rhombus, then not all four sides would be congruent; at least one of \(BC\), \(CD\), or \(AD\) would differ from \(AB\). 2. Opposite sides of a parallelogram are congruent, so \(AB=CD\) and \(BC=AD\). 3. Combining these with the given \(AB=BC\) gives \(AB=BC=CD=AD\). 4. This contradicts the consequence of the assumption that at least one side differs from \(AB\). 5. Therefore, \(ABCD\) is a rhombus.

a) Assume at least one of the other three sides is not congruent to \(AB\). b) Opposite-side congruence gives \(AB=CD\) and \(BC=AD\); with \(AB=BC\), all four sides are congruent. c) This contradicts the assumption, so the parallelogram is a rhombus.

- Negate the inequality you are trying to prove. - Express the exterior angle using the two remote interior angles. - Check what the assumed inequality would force for the remaining remote angle.

1. Assume for contradiction that \(m\angle ACD\le m\angle A\). 2. By the exterior angle theorem, \(m\angle ACD=m\angle A+m\angle B\). 3. Substituting gives \(m\angle A+m\angle B\le m\angle A\), so \(m\angle B\le0^\circ\). 4. An angle of a nondegenerate triangle must have positive measure, so this is impossible. 5. Therefore, \(m\angle ACD>m\angle A\).

Assuming \(m\angle ACD\le m\angle A\) forces \(m\angle B\le0^\circ\), which is impossible. Hence \(m\angle ACD>m\angle A\).

- Opposite angles of a cyclic quadrilateral are supplementary. - Equal supplementary angles have one forced measure. - Apply the same cyclic-angle property to the second opposite pair after finding \(m\angle B\).

1. Opposite angles in a cyclic quadrilateral are supplementary, so \(m\angle A+m\angle C=180^\circ\). 2. Since \(\angle A\cong\angle C\), each measures \(90^\circ\). The assumed congruence then also gives \(m\angle B=90^\circ\). 3. The other opposite pair is supplementary, so \(m\angle B+m\angle D=180^\circ\). Hence \(m\angle D=90^\circ\). 4. All four angles are therefore congruent, contradicting the assumption that exactly three are congruent. 5. Thus a cyclic quadrilateral cannot have exactly three congruent interior angles.

a) \(m\angle A=m\angle C=90^\circ\). b) \(m\angle D=90^\circ\). c) Then all four angles are congruent, contradicting the assumption that exactly three are congruent.

- First identify a line through the point that is tangent to the circle. - For uniqueness, assume two distinct tangent lines exist at the same point. - Relate each tangent to the radius and use uniqueness of a perpendicular through a point.

1. Let \(T\) be a point on a circle with center \(O\). The line through \(T\) perpendicular to \(\overline{OT}\) is tangent to the circle at \(T\), so a tangent at \(T\) exists. 2. To prove uniqueness, assume for contradiction that two distinct lines \(t_1\) and \(t_2\) are tangent at \(T\). 3. A tangent at \(T\) is perpendicular to radius \(\overline{OT}\), so both \(t_1\perp OT\) and \(t_2\perp OT\). 4. Through a given point, there is exactly one line perpendicular to a given line. 5. Therefore, \(t_1=t_2\), contradicting the assumption that they are distinct. 6. Hence there is exactly one tangent line at \(T\).

The line through \(T\) perpendicular to \(\overline{OT}\) is tangent to the circle, so a tangent exists. Two alleged tangents at \(T\) would both be perpendicular to \(\overline{OT}\). Since only one perpendicular to \(\overline{OT}\) passes through \(T\), the two lines would coincide. Therefore, the tangent is unique.

- Turn \(AB\ne AC\) into one of two strict length inequalities. - Write a Pythagorean equation for each right triangle formed by the altitude. - Use the midpoint condition to compare the two hypotenuse squares with the assumed inequality.

1. Assume for contradiction that \(AB\ne AC\). Then either \(AB>AC\) or \(AC>AB\). 2. Since \(\overline{AD}\) is an altitude, triangles \(ADB\) and \(ADC\) are right triangles. 3. Since \(D\) is the midpoint of \(\overline{BC}\), \(BD=DC\), and the triangles share leg \(AD\). 4. The Pythagorean theorem gives \(AB^2=AD^2+BD^2\) and \(AC^2=AD^2+DC^2\). 5. Because \(BD=DC\), these equations give \(AB^2=AC^2\), so \(AB=AC\). 6. This contradicts both possible inequalities from the assumption. Therefore, \(AB=AC\).

Assume \(AB\ne AC\), so one of the two lengths is greater. The two right triangles share leg \(AD\) and have \(BD=DC\). Thus \(AB^2=AD^2+BD^2=AD^2+DC^2=AC^2\), which forces \(AB=AC\), contradicting the assumption. Therefore, \(AB=AC\).

- Assume the diagonals do bisect each other. - Identify the quadrilateral test triggered by that diagonal property. - Compare the resulting number of parallel side pairs with the definition given.

1. Assume for contradiction that a trapezoid with exactly one pair of parallel opposite sides has diagonals that bisect each other. 2. A quadrilateral whose diagonals bisect each other is a parallelogram. 3. A parallelogram has two pairs of parallel opposite sides. 4. This contradicts the definition that the trapezoid has exactly one pair of parallel opposite sides. 5. Therefore, the diagonals of such a trapezoid cannot bisect each other.

If the diagonals bisected each other, the quadrilateral would be a parallelogram and would have two pairs of parallel sides. That contradicts the stated trapezoid definition, so its diagonals cannot bisect each other.

- Assume the nonsquare rhombus is cyclic. - Compare the opposite-angle properties of parallelograms and cyclic quadrilaterals. - Determine what congruent supplementary angles must measure.

1. Assume for contradiction that a nonsquare rhombus is cyclic. 2. Opposite angles of a rhombus are congruent because a rhombus is a parallelogram. 3. Opposite angles of a cyclic quadrilateral are supplementary. 4. Two angles that are both congruent and supplementary must each measure \(90^\circ\). 5. Thus the rhombus has right angles and is a square, contradicting the assumption that it is not a square. 6. Therefore, a nonsquare rhombus cannot be inscribed in a circle.

A cyclic rhombus would have opposite angles that are both congruent and supplementary, forcing right angles. It would therefore be a square, contradicting the nonsquare condition.

- Assume the line has two different tangency points. - Consider the radii to both alleged tangency points. - Use the uniqueness of the perpendicular from a point to a line.

1. Assume for contradiction that line \(t\) is tangent to a circle with center \(O\) at two distinct points \(A\) and \(B\). 2. A radius to a point of tangency is perpendicular to the tangent, so \(OA\perp t\) and \(OB\perp t\). 3. From point \(O\), there is exactly one perpendicular segment to line \(t\). 4. Therefore, the perpendicular feet \(A\) and \(B\) must be the same point. 5. This contradicts the assumption that \(A\) and \(B\) are distinct. Hence a line cannot be tangent to one circle at two distinct points.

Two tangency points would both be the foot of the perpendicular from the center to the same line. That perpendicular foot is unique, so the points would coincide. Therefore, two distinct tangency points are impossible.

- First intersect two internal angle bisectors and use their distance properties to establish existence. - For uniqueness, assume there are two different incenters. - Identify the same two lines on which every incenter must lie.

1. The internal angle bisectors of \(\angle A\) and \(\angle B\) intersect at a point \(I\) inside triangle \(ABC\). 2. Because \(I\) lies on the bisector of \(\angle A\), its perpendicular distances to lines \(AB\) and \(AC\) are equal. Because \(I\) lies on the bisector of \(\angle B\), its distances to lines \(BA\) and \(BC\) are equal. 3. Thus \(I\) is equidistant from all three sides, so an incenter exists. 4. To prove uniqueness, assume for contradiction that the triangle has two distinct incenters, \(I\) and \(J\). 5. Every incenter lies on the internal angle bisectors of \(\angle A\) and \(\angle B\), so both \(I\) and \(J\) lie at the intersection of those same two lines. 6. Two nonparallel lines intersect at exactly one point, so \(I=J\), contradicting the assumption that the incenters are distinct. 7. Hence a triangle has exactly one incenter.

The internal angle bisectors of \(\angle A\) and \(\angle B\) intersect at a point equidistant from all three sides, so an incenter exists. Any incenter must lie at that same unique intersection, so two distinct incenters are impossible.

- Write the side line perpendicular to each altitude. - Recall what happens to two lines perpendicular to parallel lines. - Compare that conclusion with the fact that two triangle sides meet at a vertex.

1. Assume for contradiction that the altitudes from vertices \(A\) and \(B\) are parallel. 2. The altitude from \(A\) is perpendicular to line \(BC\), and the altitude from \(B\) is perpendicular to line \(AC\). 3. If the two altitudes are parallel, then lines \(BC\) and \(AC\), each perpendicular to one of those parallel lines, must be parallel. 4. But lines \(BC\) and \(AC\) meet at vertex \(C\), so they cannot be distinct parallel lines. 5. This contradiction shows that the two altitudes cannot be parallel.

Parallel altitudes would force the two opposite side lines to be parallel, even though they meet at a triangle vertex. Therefore, two triangle altitudes cannot be parallel.

- Assume that two distinct perpendicular bisectors exist. - Identify the point through which both lines must pass. - Use the uniqueness of a perpendicular through a point on a line.

1. A segment has one midpoint \(M\), and the line through \(M\) perpendicular to the segment is a perpendicular bisector. 2. Assume for contradiction that the segment has two distinct perpendicular bisectors, \(p\) and \(q\). 3. Both \(p\) and \(q\) must pass through the same midpoint \(M\), and both must be perpendicular to the line containing the segment. 4. Through a given point on a line, only one line can be perpendicular to that line. Therefore, \(p=q\), contradicting the assumption that they are distinct. 5. Hence a nonzero segment has exactly one perpendicular bisector.

Two supposed perpendicular bisectors would both pass through the segment's unique midpoint and be perpendicular to the same line. The perpendicular through that midpoint is unique, so the two lines would be identical. Therefore, the segment has exactly one perpendicular bisector.

- Consider the triangle formed by the chord endpoints and the center. - Assume the chord exceeds the diameter. - Apply the triangle inequality to the two radii and the chord.

1. Let \(\overline{AB}\) be any chord of a circle with center \(O\) and radius \(r\). 2. Assume for contradiction that \(AB>2r\), the length of a diameter. 3. In triangle \(AOB\), \(OA=OB=r\). The triangle inequality gives \(AB\le OA+OB=2r\). 4. This contradicts the assumption \(AB>2r\). 5. Therefore, no chord can be longer than a diameter. Equality occurs only when \(A\), \(O\), and \(B\) are collinear, so the chord itself is a diameter.

A chord longer than \(2r\) would violate the triangle inequality in \(\triangle AOB\), whose other two sides are radii. Therefore, every chord has length at most the diameter.

- Assume the incenter lies on one side. - Translate that location into a perpendicular distance. - Use the defining equal-distance property of the incenter.

1. Assume for contradiction that the incenter \(I\) of triangle \(ABC\) lies on side \(\overline{AB}\). 2. The perpendicular distance from \(I\) to line \(AB\) is then \(0\). 3. An incenter is equidistant from all three side lines, so the distances from \(I\) to lines \(AC\) and \(BC\) must also be \(0\). 4. Therefore, \(I\) would lie on \(AB\), \(AC\), and \(BC\) at the same time. 5. In a nondegenerate triangle, the three side lines do not pass through one common point. This is a contradiction. 6. Hence the incenter cannot lie on a side of a nondegenerate triangle.

If the incenter lay on a side, its distance to that side would be \(0\). Equal distances to all three sides would then force it onto all three side lines, which is impossible for a nondegenerate triangle.

- Translate “inside the circle” into a comparison involving the radius. - Use the radius at the alleged tangency point to identify the distance from the center to the line. - Compare that shortest distance with \(OP\), since \(P\) lies on the line.

1. Let the circle have center \(O\) and radius \(r\), and let \(P\) be inside it, so \(OP<r\). 2. Assume for contradiction that a line \(\ell\) through \(P\) is tangent to the circle at \(T\). 3. A radius to a point of tangency is perpendicular to the tangent line, so \(OT\perp\ell\). Therefore, \(OT\) is the perpendicular distance from \(O\) to \(\ell\), and \(OT=r\). 4. Because \(P\) lies on \(\ell\), the shortest distance from \(O\) to \(\ell\) cannot exceed \(OP\). Thus \(r=OT\le OP\). 5. This contradicts \(OP<r\). Therefore, no tangent line through an interior point exists.

A supposed tangent through \(P\) would be at perpendicular distance \(r\) from the center. Since \(P\) lies on that line, its distance from the center would satisfy \(OP\ge r\), contradicting \(OP<r\).

- Assume that one point and its translated image have identical coordinates. - Compare each coordinate before and after the translation. - Use the translation vector to find the displacement length.

1. Assume for contradiction that some point \(P(x, y)\) is fixed by the translation. 2. The image of \(P\) is \(P''(x+5, y-2)\). 3. If \(P\) were fixed, then \(P''=P\), so \(x+5=x\) and \(y-2=y\). 4. These equations would require \(5=0\) and \(-2=0\), which is impossible. 5. Therefore, the translation has no fixed point. 6. The displacement length is \(\sqrt{5^2+(-2)^2}=\sqrt{29}\), so every point moves \(\sqrt{29}\) units.

The translation has no fixed point, and every point moves \(\sqrt{29}\) units.

- Ask what an “at most one” claim requires a proof to rule out. - Check whether each plan establishes existence, uniqueness, or an unrelated fact. - What would the negation of the uniqueness claim look like at the start of a contradiction proof?

1. Plan A proves existence of a perpendicular, not uniqueness. 2. Plan C does not compare two possible perpendiculars, so it does not establish “at most one.” 3. Plan B assumes the negation of uniqueness: two distinct perpendiculars through \(P\) exist. 4. Showing that those two lines must coincide contradicts the assumption that they are distinct, so Plan B is a proof by contradiction.

Plan B; proof by contradiction.

- Negate the stated conclusion, not a stronger statement that would imply it. - Ask whether an isosceles but nonequilateral triangle could satisfy \(AB=AC\). - Write the direct symbolic negation of the equality.

1. The conclusion is only \(AB=AC\). 2. “The triangle is not equilateral” does not negate that conclusion; an isosceles triangle can satisfy \(AB=AC\) without being equilateral. 3. The exact negation of \(AB=AC\) is \(AB\ne AC\). 4. Therefore, the indirect proof should begin by assuming \(AB\ne AC\).

“Triangle \(ABC\) is not equilateral” is too strong and does not negate \(AB=AC\). The correct assumption is \(AB\ne AC\).

- Use the fact that both side lines share vertex \(A\). - Distinct parallel lines do not meet. - Determine what coincident side lines would imply about the three vertices.

1. Assume for contradiction that \(\overline{AB}\parallel\overline{AC}\). 2. Both side lines pass through the same point \(A\). 3. Distinct parallel lines cannot intersect, so two lines through \(A\) cannot be distinct and parallel. 4. Therefore, the lines containing \(\overline{AB}\) and \(\overline{AC}\) would have to be the same line. 5. Then \(A\), \(B\), and \(C\) would be collinear, contradicting that \(ABC\) is a nondegenerate triangle. 6. Hence \(\overline{AB}\) and \(\overline{AC}\) cannot be parallel.

If \(AB\parallel AC\) even though both lines pass through \(A\), the two lines must coincide. That makes \(A,B,C\) collinear, contradicting that the triangle is nondegenerate.

- Replace Nuria’s claimed relationship by comparing each triangle side with the same line. - Through one point, how many lines can be perpendicular to a fixed line? - The contradiction should violate the nondegenerate-triangle condition.

1. Perpendicularity to two parallel lines does not make the two transversals perpendicular to each other. Nuria’s conclusion \(AB\perp AC\) is unsupported. 2. Since \(p\parallel q\) and \(AC\perp q\), line \(AC\) is also perpendicular to \(p\). Line \(AB\) is perpendicular to \(p\) as well. 3. Through point \(A\), there is only one line perpendicular to \(p\). Therefore, lines \(AB\) and \(AC\) would have to be the same line. 4. That would make \(A\), \(B\), and \(C\) collinear, contradicting the fact that \(ABC\) is a nondegenerate triangle. 5. Hence the assumption \(p\parallel q\) is false.

a) Being perpendicular to two parallel lines does not imply that \(AB\) and \(AC\) are perpendicular to each other. b) Under \(p\parallel q\), both \(AB\) and \(AC\) are perpendicular to \(p\), so they must be the same line through \(A\). c) That makes \(A,B,C\) collinear, contradicting the nondegenerate triangle.

- “Exactly one” can be separated into an existence claim and a uniqueness claim. - A uniqueness contradiction should assume two different objects both satisfy the requirement. - Ask which loci every possible circle center must lie on.

1. Constructing two perpendicular bisectors gives their intersection \(O\), which is equidistant from \(A\), \(B\), and \(C\). This establishes existence of a circle centered at \(O\) through all three points. 2. For uniqueness, assume there are two distinct circles through \(A\), \(B\), and \(C\), with distinct centers \(O_1\) and \(O_2\). 3. Any center of a circle through \(A\) and \(B\) lies on the perpendicular bisector of \(AB\); similarly, it lies on the perpendicular bisector of \(AC\). 4. Because the points are noncollinear, those two perpendicular bisectors meet at exactly one point. 5. Thus both \(O_1\) and \(O_2\) must equal that same intersection, contradicting the assumption that they are distinct.

a) The construction establishes existence. b) Assume two distinct circles through \(A,B,C\), with distinct centers. c) Both centers must lie at the unique intersection of the same two perpendicular bisectors, so they cannot be distinct.

- Assume the two bisectors meet at a right angle. - Examine the smaller triangle formed by the two vertices and the bisectors’ intersection. - Compare its angle sum with the angle sum of the original triangle.

1. In triangle \(ABC\), let the internal bisectors of \(\angle A\) and \(\angle B\) meet at \(I\). Assume for contradiction that the bisectors are perpendicular, so \(\angle AIB=90^\circ\). 2. In triangle \(AIB\), the other two angles have measures \(\frac{A}{2}\) and \(\frac{B}{2}\). 3. The triangle angle sum gives \(\frac{A}{2}+\frac{B}{2}+90^\circ=180^\circ\), so \(A+B=180^\circ\). 4. Then the angle sum of triangle \(ABC\) forces \(C=0^\circ\), which is impossible for a nondegenerate triangle. 5. Therefore, two internal angle bisectors of a nondegenerate triangle cannot be perpendicular.

Assuming the two internal bisectors are perpendicular forces the third angle of the triangle to be \(0^\circ\). This contradiction proves that the bisectors cannot be perpendicular.

- Negate the desired angle inequality carefully. - Separate the equality case from the strict-inequality case. - Relate the order of two angles to the order of their opposite sides.

1. Assume for contradiction that \(\angle C\le\angle B\). 2. If \(\angle C=\angle B\), then the opposite sides are congruent, so \(AB=AC\), contradicting \(AB>AC\). 3. If \(\angle C<\angle B\), then the side opposite \(\angle C\) is shorter than the side opposite \(\angle B\), so \(AB<AC\), again contradicting \(AB>AC\). 4. Both alternatives are impossible. Therefore, \(\angle C>\angle B\).

Assuming \(\angle C\le\angle B\) forces either \(AB=AC\) or \(AB<AC\), both of which contradict \(AB>AC\). Hence \(\angle C>\angle B\).

- Translate “not equilateral” into a statement about at least one pair of side lengths. - Apply the angle bisector theorem together with the midpoint condition for each median. - Compare the two side equalities with the consequence of the assumption.

1. Assume for contradiction that triangle \(ABC\) is not equilateral. Then at least one pair of its side lengths is unequal. 2. Let the median from \(A\) meet \(\overline{BC}\) at \(D\). Since \(BD=DC\) and \(\overline{AD}\) bisects \(\angle A\), the angle bisector theorem gives \(AB=AC\). 3. Let the median from \(B\) meet \(\overline{AC}\) at \(E\). Since \(AE=EC\) and \(\overline{BE}\) bisects \(\angle B\), the angle bisector theorem gives \(BA=BC\). 4. Therefore, \(AB=AC=BC\), so no pair of side lengths is unequal. 5. This contradicts the consequence of the assumption in step 1. Hence triangle \(ABC\) is equilateral.

Assume the triangle is not equilateral, so at least one pair of sides is unequal. The median-angle-bisector from \(A\) gives \(AB=AC\), and the one from \(B\) gives \(AB=BC\). Thus all three sides are congruent, contradicting the assumption. Therefore, the triangle is equilateral.

- Express “not equilateral” as a consequence involving unequal angle measures. - Combine the angle-bisector property of the incenter with the equal-radius property of the circumcenter. - Repeat the comparison for a second pair of vertices and compare the result with the assumption.

1. Assume for contradiction that triangle \(ABC\) has a common incenter and circumcenter \(O\) but is not equilateral. Then at least two of its angles have different measures. 2. Since \(O\) is the incenter, \(AO\) and \(BO\) bisect \(\angle A\) and \(\angle B\). 3. Since \(O\) is the circumcenter, \(OA=OB\). Thus triangle \(AOB\) is isosceles, so \(\angle OAB=\angle ABO\). 4. These angles equal \(\frac{1}{2}\angle A\) and \(\frac{1}{2}\angle B\), so \(\angle A=\angle B\). 5. Applying the same argument to another pair of vertices gives \(\angle B=\angle C\). Hence \(\angle A=\angle B=\angle C\). 6. This contradicts the consequence of the assumption that at least two angles have different measures. Therefore, the triangle is equilateral.

Assume the triangle is not equilateral, so at least two angles differ. The common center lies on every angle bisector and is equidistant from the vertices. Each center triangle is therefore isosceles, which gives \(\angle A=\angle B=\angle C\). This contradicts the assumption, so the triangle is equilateral.

- Turn \(AB\ne AC\) into one of two strict inequalities. - Multiply that inequality by the common positive altitude factor used in the area formula. - Compare the two resulting expressions as areas of the same triangle.

1. Let the common altitude length be \(h>0\). Assume for contradiction that \(AB\ne AC\). Then either \(AB>AC\) or \(AC>AB\). 2. If \(AB>AC\), multiplying by the positive number \(\frac{h}{2}\) gives \(\frac{1}{2}ABh>\frac{1}{2}ACh\). If \(AC>AB\), the reverse strict inequality follows. 3. But \(\frac{1}{2}ABh\) and \(\frac{1}{2}ACh\) are both expressions for the area of the same triangle, using the two equal altitudes. 4. The same area cannot be both greater and less than itself. Thus either inequality from the assumption is impossible. 5. Therefore, \(AB=AC\).

Assume \(AB\ne AC\), so one side is longer. Multiplying the strict inequality by the common positive factor \(\frac{h}{2}\) would make one of the area expressions \(\frac{1}{2}ABh\) and \(\frac{1}{2}ACh\) larger than the other. Both equal the area of \(\triangle ABC\), a contradiction. Therefore, \(AB=AC\).

- Negate the equality that expresses “equidistant.” - If two positive distances are unequal, name the larger one and compare the associated right triangles. - Use the Pythagorean theorem to relate each half-chord to its distance from the center.

1. The conclusion to be proved is \(OM=ON\), so the contradiction assumption is \(OM\ne ON\). 2. Without loss of generality, suppose \(OM>ON\). 3. A perpendicular from the center to a chord bisects the chord, so \(AM=\frac{AB}{2}\) and \(CN=\frac{CD}{2}\). 4. Right triangles \(\triangle OMA\) and \(\triangle ONC\) have equal hypotenuse lengths because \(OA=OC=r\). 5. By the Pythagorean theorem, \(AM^2=r^2-OM^2\) and \(CN^2=r^2-ON^2\). 6. Since \(OM>ON\), it follows that \(OM^2>ON^2\), so \(AM^2<CN^2\) and therefore \(AM<CN\). 7. Thus \(AB=2AM<2CN=CD\), contradicting the given fact that \(AB=CD\). 8. Therefore, \(OM=ON\), and the congruent chords are equidistant from the center.

a) Assume for contradiction that \(OM\ne ON\). b) If, without loss of generality, \(OM>ON\), then \(AM^2=r^2-OM^2<r^2-ON^2=CN^2\). Hence \(AB=2AM<2CN=CD\), contradicting \(AB=CD\). Therefore, \(OM=ON\).

- Let the perpendicular distances from the center to the two chords be \(OM\) and \(ON\). - Express each half-chord using a radius and the center-to-chord distance. - Assume the longer chord is not closer and compare the resulting half-lengths.

1. Let \(M\) and \(N\) be the feet of the perpendiculars from \(O\) to \(\overline{AB}\) and \(\overline{CD}\), respectively. These perpendiculars bisect the chords. 2. Assume for contradiction that \(AB\) is not closer to \(O\), so \(OM\ge ON\). 3. In right triangles \(OMA\) and \(ONC\), the hypotenuses \(OA\) and \(OC\) are equal radii. 4. From the Pythagorean theorem, \(AM^2=OA^2-OM^2\) and \(CN^2=OC^2-ON^2\). Since \(OM\ge ON\), it follows that \(AM^2\le CN^2\), so \(AM\le CN\). 5. Therefore, \(AB=2AM\le2CN=CD\), contradicting \(AB>CD\). 6. Hence \(OM<ON\), so the longer chord \(\overline{AB}\) is closer to the center.

Assuming the longer chord is at least as far from the center forces its half-length to be no greater than the other chord's half-length, contradicting \(AB>CD\). Therefore, the longer chord is closer to the center.

- Assume the median is also perpendicular to the side it meets. - Identify the special line created by a midpoint and a perpendicular. - Combine the resulting equal lengths with the given isosceles condition.

1. Let \(D\) be the midpoint of \(\overline{AC}\), so \(BD\) is the median from \(B\). 2. Assume for contradiction that \(BD\) is also an altitude. Then \(BD\perp AC\). 3. Since \(D\) is the midpoint of \(AC\) and \(BD\perp AC\), line \(BD\) is the perpendicular bisector of \(\overline{AC}\). 4. Point \(B\) lies on this perpendicular bisector, so \(BA=BC\). 5. The given condition is \(BA=AC\). Therefore, \(BA=BC=AC\), making the triangle equilateral. 6. This contradicts the statement that the triangle is not equilateral. Hence the median from \(B\) to \(AC\) cannot also be an altitude.

If the median from \(B\) were also perpendicular to \(AC\), it would be the perpendicular bisector of \(AC\), giving \(BA=BC\). Together with \(BA=AC\), this would make the triangle equilateral, a contradiction.

- Check whether each intermediate statement actually depends on the opening assumption. - A contradiction proof should use the contrary assumption to force a statement that conflicts with another established fact. - Relate the equal distances from \(M\) to a circle and compare that with what the contrary assumption says about point \(A\).

1. The argument is logically valid, but the assumption \(\angle A\ne90^\circ\) is not used to derive any intermediate fact. Deleting the assumption and the final contradiction leaves a complete direct proof. 2. For a genuine indirect proof, assume \(\angle A\ne90^\circ\). 3. By the contrapositive of the theorem that an angle subtending a diameter is a right angle, point \(A\) cannot lie on the circle with diameter \(\overline{BC}\). 4. Since \(M\) is the midpoint of \(BC\), \(BM=CM=\frac{1}{2}BC\). The given condition also gives \(AM=\frac{1}{2}BC\). 5. Therefore, \(AM=BM=CM\), so \(A\), \(B\), and \(C\) lie on the circle centered at \(M\). 6. Because \(M\) lies on \(BC\), segment \(\overline{BC}\) is a diameter of that circle. Thus \(A\) does lie on the circle with diameter \(\overline{BC}\), contradicting Step 3. 7. Hence \(\angle A=90^\circ\).

a) The argument is logically valid, but it is a direct proof with an unused contradiction assumption: the circle-and-diameter reasoning proves \(\angle A=90^\circ\) without the first sentence. b) Assume \(\angle A\ne90^\circ\). Then, by the contrapositive of the diameter-angle theorem, \(A\) is not on the circle with diameter \(BC\). But \(AM=BM=CM\) places \(A\), \(B\), and \(C\) on the circle centered at \(M\), whose diameter is \(BC\). This contradiction proves \(\angle A=90^\circ\).

- Turn \(AB\ne AC\) into one strict inequality by naming the longer side. - Use the shared altitude in two Pythagorean equations to compare \(BD\) and \(CD\). - Use the equal bisected angles and the two right angles to obtain a conflicting comparison.

1. Assume for contradiction that \(AB\ne AC\). Without loss of generality, suppose \(AB>AC\). 2. Since \(AD\perp BC\), triangles \(ABD\) and \(ACD\) are right triangles sharing leg \(AD\). 3. By the Pythagorean theorem, \(BD^2=AB^2-AD^2\) and \(CD^2=AC^2-AD^2\). 4. The assumption \(AB>AC\) gives \(AB^2>AC^2\), so \(BD^2>CD^2\) and therefore \(BD>CD\). 5. However, \(\angle BAD=\angle DAC\) because \(AD\) bisects \(\angle A\), and \(\angle ADB=\angle ADC=90^\circ\). Thus \(\triangle ABD\sim\triangle ACD\) by AA. 6. Side \(AD\) corresponds to itself, so the similarity scale factor is \(1\). Therefore, \(BD=CD\). 7. This contradicts \(BD>CD\). Hence \(AB=AC\).

Assume, without loss of generality, that \(AB>AC\). The Pythagorean theorem then gives \(BD^2=AB^2-AD^2>AC^2-AD^2=CD^2\), so \(BD>CD\). But the two right triangles are AA-similar, and their shared corresponding side \(AD\) makes the scale factor \(1\), forcing \(BD=CD\). This contradiction proves \(AB=AC\).

- First rule out coincident centers using one shared point. - Apply the center-tangency-point alignment at both alleged tangency points. - Determine what two shared points on the line of centers mean for both circles.

1. Assume for contradiction that distinct circles with centers \(O_1\) and \(O_2\) are tangent at two different points \(A\) and \(B\). 2. The centers must be distinct. If \(O_1=O_2\), the shared point \(A\) would give both circles the same radius, so the circles would be identical. 3. At a point of tangency, the two centers and the tangency point are collinear. Therefore, \(O_1\), \(O_2\), and \(A\) are collinear, and \(O_1\), \(O_2\), and \(B\) are collinear. 4. Thus, both \(A\) and \(B\) lie on line \(O_1O_2\). 5. For either circle, two distinct points on the circle and on a line through its center are the endpoints of a diameter. Hence \(A\) and \(B\) are diameter endpoints for both circles. 6. Both circles therefore have the same midpoint of \(\overline{AB}\) as center and the same radius \(\frac{1}{2}AB\), so they are the same circle. 7. This contradicts the assumption that the circles are distinct. Therefore, two distinct circles cannot be tangent at two different points.

Assuming two tangency points forces both circles to have the same diameter \(\overline{AB}\), hence the same center and radius. That contradicts their being distinct.

- Form three smaller triangles using the interior point. - Write one triangle inequality for each side of the original triangle. - Add the inequalities and compare the result with the assumption.

1. Assume for contradiction that \(PA+PB+PC\le\frac{1}{2}(AB+BC+CA)\). 2. The triangle inequalities in triangles \(APB\), \(BPC\), and \(CPA\) give \(AB<PA+PB\), \(BC<PB+PC\), and \(CA<PC+PA\). 3. Adding these inequalities gives \(AB+BC+CA<2(PA+PB+PC)\). 4. Dividing by \(2\) gives \(\frac{1}{2}(AB+BC+CA)<PA+PB+PC\). 5. This contradicts the assumed inequality. Therefore, the stated strict inequality holds.

The three triangle inequalities add to \(AB+BC+CA<2(PA+PB+PC)\), contradicting the assumption that the distance sum is at most half the perimeter.

- Double the median by reflecting its endpoint across the side midpoint. - Identify the parallelogram created by the two bisecting diagonals. - Apply the triangle inequality to the new triangle.

1. Assume for contradiction that \(m_a\ge\frac{AB+AC}{2}\). 2. Reflect \(A\) across \(M\) to point \(A''\). Then \(AA''=2m_a\). 3. Since \(M\) is the midpoint of both \(\overline{BC}\) and \(\overline{AA''}\), quadrilateral \(ABA''C\) is a parallelogram. 4. Therefore, \(BA''=AC\). 5. In triangle \(ABA''\), the triangle inequality gives \(AA''<AB+BA''=AB+AC\). 6. Thus, \(2m_a<AB+AC\), or \(m_a<\frac{AB+AC}{2}\), contradicting the assumption.

Reflecting \(A\) across the midpoint of \(BC\) creates a triangle in which the triangle inequality gives \(2m_a<AB+AC\). Therefore, \(m_a<\frac{AB+AC}{2}\).

- Assume three distinct intersection points occur on one line. - Use equal radii to place the center on perpendicular-bisector loci. - Compare the two bisectors' directions and their intersection points with the original line.

1. Assume for contradiction that line \(\ell\) intersects a circle with center \(O\) at three distinct points \(A\), \(B\), and \(C\), in that order. 2. Since \(OA=OB\), point \(O\) lies on the perpendicular bisector \(p\) of \(\overline{AB}\). 3. Since \(OB=OC\), point \(O\) lies on the perpendicular bisector \(q\) of \(\overline{BC}\). 4. Segments \(\overline{AB}\) and \(\overline{BC}\) lie on the same line \(\ell\), so both \(p\) and \(q\) are perpendicular to \(\ell\). 5. Because \(p\) and \(q\) both pass through \(O\), the uniqueness of a perpendicular through a point implies that they are the same line. 6. That one line would meet \(\ell\) at both the midpoint of \(\overline{AB}\) and the midpoint of \(\overline{BC}\). These midpoints are distinct because \(A\), \(B\), and \(C\) are distinct and ordered on \(\ell\). 7. This is impossible because two distinct lines intersect at most once. Therefore, a line intersects a circle at no more than two points.

Three intersections would place the circle's center on the perpendicular bisectors of two collinear subsegments. Both bisectors pass through the center and are perpendicular to the original line, so uniqueness forces them to be the same line. That line would then intersect the original line at two distinct midpoints, which is impossible. Therefore, at most two intersections can occur.

- Assume all three median comparisons fail at once. - Convert each comparison using a median-length formula. - Add the resulting three side-length inequalities.

1. Let the side lengths be \(a\), \(b\), and \(c\), with corresponding medians \(m_a\), \(m_b\), and \(m_c\). 2. Assume for contradiction that \(m_a\le\frac{a}{2}\), \(m_b\le\frac{b}{2}\), and \(m_c\le\frac{c}{2}\). 3. The median formulas give \(4m_a^2=2b^2+2c^2-a^2\), with analogous equations for the other two medians. 4. From \(m_a\le\frac{a}{2}\), obtain \(b^2+c^2\le a^2\). Similarly, \(c^2+a^2\le b^2\) and \(a^2+b^2\le c^2\). 5. Adding the three inequalities gives \(2(a^2+b^2+c^2)\le a^2+b^2+c^2\). 6. This would force \(a^2+b^2+c^2\le0\), impossible for a nondegenerate triangle. Therefore, at least one median is longer than half its corresponding side.

Assuming all three medians are at most half their corresponding sides produces the impossible inequality \(2(a^2+b^2+c^2)\le a^2+b^2+c^2\). Hence at least one median is longer.

- Assume two side lengths are unequal and order them from shorter to longer. - Compare the squared lengths of the medians to those two sides. - Look for a positive difference that contradicts congruent medians.

1. Assume for contradiction that triangle \(ABC\) has three congruent medians but is not equilateral. 2. Let side \(c\) be longer than side \(a\). Let \(m_a\) and \(m_c\) be the medians to sides \(a\) and \(c\). 3. By the median-length formulas, \(m_a^2=\frac{2b^2+2c^2-a^2}{4}\) and \(m_c^2=\frac{2a^2+2b^2-c^2}{4}\). 4. Subtracting gives \(m_a^2-m_c^2=\frac{3(c^2-a^2)}{4}>0\). 5. Thus, \(m_a>m_c\), contradicting the assumption that all three medians are congruent. 6. Therefore, no two side lengths can differ, and the triangle is equilateral.

If one side were longer than another, the median to the shorter side would be longer than the median to the longer side. This contradicts three congruent medians, so all three sides are congruent.

- The common point is inside the triangle because it is the incenter; use this to identify the triangle as acute. - Recall the angle formed at an incenter by the bisectors from \(B\) and \(C\), and the angle formed at an orthocenter by the altitudes from those vertices. - Compare both formulas with \(120^\circ\) after assuming \(\angle A\) is greater than or less than \(60^\circ\).

1. Let the common orthocenter and incenter be \(P\). Because an incenter lies inside its triangle, \(P\) is inside the triangle, so the triangle is acute. 2. Assume for contradiction that the triangle is not equilateral. Then at least one angle is not \(60^\circ\); relabel so \(\angle A\ne60^\circ\). 3. Since \(P\) is the incenter, \(m\angle BPC=90^\circ+\frac{1}{2}m\angle A\). 4. Since \(P\) is also the orthocenter of an acute triangle, \(m\angle BPC=180^\circ-m\angle A\). 5. If \(m\angle A>60^\circ\), the incenter formula makes \(m\angle BPC>120^\circ\), while the orthocenter formula makes \(m\angle BPC<120^\circ\), which is impossible. 6. If \(m\angle A<60^\circ\), the incenter formula makes \(m\angle BPC<120^\circ\), while the orthocenter formula makes \(m\angle BPC>120^\circ\), which is also impossible. 7. Both alternatives contradict the assumption \(\angle A\ne60^\circ\). Therefore, every angle is \(60^\circ\), and the triangle is equilateral.

Assume some angle, say \(\angle A\), is not \(60^\circ\). The incenter identity gives \(m\angle BPC=90^\circ+\frac{1}{2}m\angle A\), while the orthocenter identity gives \(m\angle BPC=180^\circ-m\angle A\). If \(\angle A\) is greater than \(60^\circ\), these values lie on opposite sides of \(120^\circ\); the same happens in reverse if it is less than \(60^\circ\). This contradiction forces all angles to be \(60^\circ\).