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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 \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.
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==Problem==
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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>
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==Solution==
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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.
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'''Lemma.''' <math>XY = \frac{r_1 + r_2}{d} \sqrt{d^2 - (r_1 - r_2)^2}.</math>
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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.
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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.
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Lemma 2. Triangles <math>O_1 A O_2</math> and <math>BAC</math> are similar.
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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.
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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>:
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<cmath>XY = \frac{AB + AC}{BC} \cdot \frac{AP}{2 h_a} \sqrt{BC^2 - (AB - AC)^2},</cmath>
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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
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<cmath>\left( \frac{2a h_a}{(b+c) \sqrt{a^2 - (b-c)^2}} \right)^2 + \frac{PB \cdot PC}{bc} = 1.</cmath>
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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
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<cmath>\frac{(a + b + c)(-a + b + c)}{(b + c)^2} + \frac{PB \cdot PC}{bc} = 1,</cmath>
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which by <math>(a + b + c)(-a + b + c) = (b + c)^2 - a^2</math> becomes
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<cmath>\frac{PB \cdot PC}{bc} = \frac{a^2 bc}{(b+c)^2}.</cmath>
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Let <math>PB = x</math>; then <math>PC = a - x</math>. The quadratic in <math>x</math> is
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<cmath>x^2 - ax + \frac{a^2 bc}{(b+c)^2} = 0,</cmath>
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which factors as
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<cmath>\left(x - \frac{ab}{b+c}\right)\left(x - \frac{ac}{b+c}\right) = 0.</cmath>
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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.
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{{MAA Notice}}

Latest revision as of 17:29, 10 May 2023

Problem

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

Solution

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

Lemma. $XY = \frac{r_1 + r_2}{d} \sqrt{d^2 - (r_1 - r_2)^2}.$

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

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

Lemma 2. Triangles $O_1 A O_2$ and $BAC$ are similar.

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

Also, let $O_1 O_2$ intersect $AP$ at $Z$. Then obviously $Z$ is the midpoint of $AP$ and $AZ$ is an altitude of triangle $A O_1 O_2$.Thus, we can simplify our expression of $XY$: \[XY = \frac{AB + AC}{BC} \cdot \frac{AP}{2 h_a} \sqrt{BC^2 - (AB - AC)^2},\] where $h_a$ is the length of the altitude from $A$ in triangle $ABC$. Hence, substituting into our condition and using $AB = c, BC = a, CA = b$ gives \[\left( \frac{2a h_a}{(b+c) \sqrt{a^2 - (b-c)^2}} \right)^2 + \frac{PB \cdot PC}{bc} = 1.\] Using $2 a h_a = 4[ABC] = \sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}$ by Heron's Formula (where $[ABC]$ is the area of triangle $ABC$, our condition becomes \[\frac{(a + b + c)(-a + b + c)}{(b + c)^2} + \frac{PB \cdot PC}{bc} = 1,\] which by $(a + b + c)(-a + b + c) = (b + c)^2 - a^2$ becomes \[\frac{PB \cdot PC}{bc} = \frac{a^2 bc}{(b+c)^2}.\] Let $PB = x$; then $PC = a - x$. The quadratic in $x$ is \[x^2 - ax + \frac{a^2 bc}{(b+c)^2} = 0,\] which factors as \[\left(x - \frac{ab}{b+c}\right)\left(x - \frac{ac}{b+c}\right) = 0.\] Hence, $PB = \frac{ab}{b+c}$ or $\frac{ac}{b+c}$, and so the $P$ corresponding to these lengths are our answer.

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