Question

Sea $\triangle ABC$ con longitudes de lado $AB = 5,$ $BC = 9,$ y $CA = 10.$ Las tangentes al circuncírculo del $\triangle ABC$ en $B$ y $C$ se intersecan en el punto $D,$ y $\overline{AD}$ interseca al circuncírculo en $P \ne A.$ La longitud de $AP$ es igual a $\frac{m}{n},$ donde $m$ y $n$ son enteros primos entre sí. Halla $m + n.$

Let $\triangle ABC$ have side lengths $AB = 5,$ $BC = 9,$ and $CA = 10.$ The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ intersect at point $D,$ and $\overline{AD}$ intersects the circumcircle at $P \ne A.$ The length of $AP$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime integers. Find $m + n.$

Respuesta

Solución:

Por el ángulo tangente-cuerda, $\angle DBC = \angle DCB = \angle A,$ así que el triángulo $DBC$ es isósceles con $DB = \frac{BC/2}{\cos A}.$ La ley de cosenos da $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ y $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15},$ así que $DB = \frac{9/2}{11/25} = \frac{225}{22}.$

Como $D$ está en el lado opuesto de $BC$ respecto de $A,$ $\angle ABD = A + B,$ y $\begin{aligned} &\cos(A + B) \\ &= \frac{11}{25} \cdot \frac{1}{15} \\ &\quad {}- \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} \\ &= \frac{11 - 336}{375} \\ &= -\frac{13}{15}. \end{aligned}$ La ley de cosenos en el triángulo $ABD$ da entonces $\begin{aligned} &DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 \\ &\quad {}+ 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} \\ &= \frac{105625}{484}, \end{aligned}$ $DA = \frac{325}{22}.$

La potencia de $D$ da $DP \cdot DA = DB^2,$ así que $\begin{aligned} &AP = DA - \frac{DB^2}{DA} \\ &= \frac{DA^2 - DB^2}{DA} \\ &= \frac{(105625 - 50625)/484}{325/22} \\ &= \frac{55000}{22 \cdot 325} \\ &= \frac{100}{13}, \end{aligned}$ y $m + n = 100 + 13 = 113.$

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so triangle $DBC$ is isosceles with $DB = \frac{BC/2}{\cos A}.$ The law of cosines gives $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ and $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15},$ so $DB = \frac{9/2}{11/25} = \frac{225}{22}.$

Since $D$ lies on the opposite side of $BC$ from $A,$ $\angle ABD = A + B,$ and $\begin{aligned} &\cos(A + B) \\ &= \frac{11}{25} \cdot \frac{1}{15} \\ &\quad {}- \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} \\ &= \frac{11 - 336}{375} \\ &= -\frac{13}{15}. \end{aligned}$ The law of cosines in triangle $ABD$ then gives $\begin{aligned} &DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 \\ &\quad {}+ 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} \\ &= \frac{105625}{484}, \end{aligned}$ $DA = \frac{325}{22}.$

The power of $D$ gives $DP \cdot DA = DB^2,$ so $\begin{aligned} &AP = DA - \frac{DB^2}{DA} \\ &= \frac{DA^2 - DB^2}{DA} \\ &= \frac{(105625 - 50625)/484}{325/22} \\ &= \frac{55000}{22 \cdot 325} \\ &= \frac{100}{13}, \end{aligned}$ and $m + n = 100 + 13 = 113.$

2024 AIME I Problema 10

Solución:

El Problema 10 en otros años