# Difference between revisions of "2013 USAMO Problems/Problem 6"

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− | Let <math>ABC</math> be a triangle. Find all points <math>P</math> on segment <math>BC</math> satisfying the following property: If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1</cmath>. | + | ==Problem== |

+ | Let <math>ABC</math> be a triangle. Find all points <math>P</math> on segment <math>BC</math> satisfying the following property: If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath> | ||

+ | |||

+ | ==Solution== | ||

+ | Let circle <math>PAB</math> (i.e. the circumcircle of <math>PAB</math>), <math>PAC</math> be <math>\omega_1, \omega_2</math> with radii <math>r_1</math>, <math>r_2</math> and centers <math>O_1, O_2</math>, respectively, and <math>d</math> be the distance between their centers. | ||

+ | |||

+ | '''Lemma.''' <math>XY = \frac{r_1 + r_2}{d} \sqrt{d^2 - (r_1 - r_2)^2}.</math> | ||

+ | |||

+ | Proof. Let the external tangent containing <math>X</math> meet <math>\omega_1</math> at <math>X_1</math> and <math>\omega_2</math> at <math>X_2</math>, and let the external tangent containing <math>Y</math> meet <math>\omega_1</math> at <math>Y_1</math> and <math>\omega_2</math> at <math>Y_2</math>. Then clearly <math>X_1 Y_1</math> and <math>X_2 Y_2</math> are parallel (for they are both perpendicular <math>O_1 O_2</math>), and so <math>X_1 Y_1 Y_2 X_2</math> is a trapezoid. | ||

+ | |||

+ | Now, <math>X_1 X^2 = XA \cdot XP = X_2 X^2</math> by Power of a Point, and so <math>X</math> is the midpoint of <math>X_1 X_2</math>. Similarly, <math>Y</math> is the midpoint of <math>Y_1 Y_2</math>. Hence, <math>XY = \frac{1}{2} (X_1 Y_1 + X_2 Y_2).</math> Let <math>X_1 Y_1</math>, <math>X_2 Y_2</math> meet <math>O_1 O_2</math> s at <math>Z_1, Z_2</math>, respectively. Then by similar triangles and the Pythagorean Theorem we deduce that <math>X_1 Z_1 = \frac{r_1 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math> and <math>\frac{r_2 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math>. But it is clear that <math>Z_1</math>, <math>Z_2</math> is the midpoint of <math>X_1 Y_1</math>, <math>X_2 Y_2</math>, respectively, so <math>XY = \frac{(r_1 + r_2)}{d} \sqrt{d^2 - (r_1 - r_2)^2},</math> as desired. | ||

+ | |||

+ | Lemma 2. Triangles <math>O_1 A O_2</math> and <math>BAC</math> are similar. | ||

+ | |||

+ | Proof. <math>\angle{AO_1 O_2} = \frac{\angle{PO_1 A}}{2} = \angle{ABC}</math> and similarly <math>\angle{AO_2 O_1} = \angle{ACB}</math>, so the triangles are similar by AA Similarity. | ||

+ | |||

+ | Also, let <math>O_1 O_2</math> intersect <math>AP</math> at <math>Z</math>. Then obviously <math>Z</math> is the midpoint of <math>AP</math> and <math>AZ</math> is an altitude of triangle <math>A O_1 O_2</math>.Thus, we can simplify our expression of <math>XY</math>: | ||

+ | <cmath>XY = \frac{AB + AC}{BC} \cdot \frac{AP}{2 h_a} \sqrt{BC^2 - (AB - AC)^2},</cmath> | ||

+ | where <math>h_a</math> is the length of the altitude from <math>A</math> in triangle <math>ABC</math>. Hence, substituting into our condition and using <math>AB = c, BC = a, CA = b</math> gives | ||

+ | <cmath>\left( \frac{2a h_a}{(b+c) \sqrt{a^2 - (b-c)^2}} \right)^2 + \frac{PB \cdot PC}{bc} = 1.</cmath> | ||

+ | Using <math>2 a h_a = 4[ABC] = \sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}</math> by Heron's Formula (where <math>[ABC]</math> is the area of triangle <math>ABC</math>, our condition becomes | ||

+ | <cmath>\frac{(a + b + c)(-a + b + c)}{(b + c)^2} + \frac{PB \cdot PC}{bc} = 1,</cmath> | ||

+ | which by <math>(a + b + c)(-a + b + c) = (b + c)^2 - a^2</math> becomes | ||

+ | <cmath>\frac{PB \cdot PC}{bc} = \frac{a^2 bc}{(b+c)^2}.</cmath> | ||

+ | Let <math>PB = x</math>; then <math>PC = a - x</math>. The quadratic in <math>x</math> is | ||

+ | <cmath>x^2 - ax + \frac{a^2 bc}{(b+c)^2} = 0,</cmath> | ||

+ | which factors as | ||

+ | <cmath>(x - \frac{ab}{b+c})(x - \frac{ac}{b+c}) = 0.</cmath> | ||

+ | Hence, <math>PB = \frac{ab}{b+c}</math> or <math>\frac{ac}{b+c}</math>, and so the <math>P</math> corresponding to these lengths are our answer. | ||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 22:54, 20 May 2015

## Problem

Let be a triangle. Find all points on segment satisfying the following property: If and are the intersections of line with the common external tangent lines of the circumcircles of triangles and , then

## Solution

Let circle (i.e. the circumcircle of ), be with radii , and centers , respectively, and be the distance between their centers.

**Lemma.**

Proof. Let the external tangent containing meet at and at , and let the external tangent containing meet at and at . Then clearly and are parallel (for they are both perpendicular ), and so is a trapezoid.

Now, by Power of a Point, and so is the midpoint of . Similarly, is the midpoint of . Hence, Let , meet s at , respectively. Then by similar triangles and the Pythagorean Theorem we deduce that and . But it is clear that , is the midpoint of , , respectively, so as desired.

Lemma 2. Triangles and are similar.

Proof. and similarly , so the triangles are similar by AA Similarity.

Also, let intersect at . Then obviously is the midpoint of and is an altitude of triangle .Thus, we can simplify our expression of : where is the length of the altitude from in triangle . Hence, substituting into our condition and using gives Using by Heron's Formula (where is the area of triangle , our condition becomes which by becomes Let ; then . The quadratic in is which factors as Hence, or , and so the corresponding to these lengths are our answer. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.