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Problem 1761

National Olympiad, first round

Geometry Difficulty 6.1 Prove it HMMT February · United States

Let $ABC$ be a triangle such that $AB = 6$ , $BC = 5$ , $AC = 7$ . Let the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at $X$ . Let $Z$ be a point on the circumcircle of $ABC$ . Let $Y$ be the foot of the perpendicular from $X$ to $CZ$ . Let $K$ be the intersection of the circumcircle of $BCY$ with line $AB$ . Given that $Y$ is on the interior of segment $CZ$ and $YZ = 3 \, CY$ , compute $AK$ .

Proposed by: Allen Liu

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Let $\omega_1$ denote the circumcircle of $ABC$ and $\omega_2$ denote the circle centered at $X$ through $B$ and $C$ . Let $\omega_2$ intersect $AB$ , $AC$ again at $B'$ , $C'$ . The (signed) power of $Y$ with respect to $\omega_1$ is $-CY \cdot YZ$ . The power of $Y$ with respect to $\omega_2$ is $XY^2 - CX^2 = -CY^2$ . Thus the ratio of the powers of $Y$ with respect to the two circles is $3 : 1$ . The circumcircle of $BCY$ passes through the intersection points of $\omega_1$ and $\omega_2$ ( $B$ and $C$ ) and thus contains exactly the set of points such that the ratio of their powers with respect to $\omega_1$ and $\omega_2$ is $3 : 1$ (this fact can be verified in a variety of ways). We conclude that $K$ must be the point on line $AB$ such that $\frac{KB'}{KA} = 3$ . It now suffices to compute $AB'$ . Note $AB' = \frac{AC \cdot B'C'}{BC}$ by similar triangles. Also an angle chase gives that $B'XC'$ are collinear. We compute

$BX = \frac{BC}{2 \cos A} = \frac{5}{2 \cdot \frac{5}{7}} = \frac{7}{2}$
and thus $B'C' = 7$ so $AB' = \frac{49}{5}$ and $AK = \frac{3}{2} AB' = \frac{147}{10}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.