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2013 USAMO Problems/Problem 6


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 \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]


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. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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