Question

Sea $\triangle ABC$ un triángulo con $D$ en $\overline{BC}$ tal que $\overline{AD}$ biseca $\angle BAC.$ Sea $\omega$ la circunferencia que pasa por $A$ y es tangente al segmento $\overline{BC}$ en $D.$ Sean $E \ne A$ y $F \ne A$ las intersecciones de $\omega$ con los segmentos $\overline{AB}$ y $\overline{AC},$ respectivamente. Supongamos que $AB = 200,$ $AC = 225,$ y que todos $AE,$ $AF,$ $BD,$ y $CD$ son enteros positivos. Halla el mayor valor posible de $BC.$

Let $\triangle ABC$ be a triangle with $D$ on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC.$ Let $\omega$ be the circle that passes through $A$ and is tangent to segment $\overline{BC}$ at $D.$ Let $E \ne A$ and $F \ne A$ be the intersections of $\omega$ with segments $\overline{AB}$ and $\overline{AC},$ respectively. Suppose that $AB = 200,$ $AC = 225,$ and all of $AE,$ $AF,$ $BD,$ and $CD$ are positive integers. Find the greatest possible value of $BC.$

Respuesta

Solución:

Como $\omega$ es tangente a $BC$ en $D,$ la potencia de $B$ da $BD^2 = BE \cdot BA$ y la potencia de $C$ da $CD^2 = CF \cdot CA.$ La bisectriz del ángulo da $\frac{BD}{DC} = \frac{AB}{AC} = \frac{8}{9},$ así que $BD = 8t$ y $CD = 9t,$ donde $t = CD - BD$ es un entero positivo. Entonces $\begin{aligned} &BE = \frac{64t^2}{200} = \frac{8t^2}{25}, \\ &CF = \frac{81t^2}{225} = \frac{9t^2}{25}, \end{aligned}$ así que $AE = 200 - \frac{8t^2}{25}$ y $AF = 225 - \frac{9t^2}{25}.$

Para que $AE$ y $AF$ sean enteros necesitamos $25 \mid t^2,$ es decir, $t = 5s.$ Entonces $AE = 200 - 8s^2 \gt 0$ obliga a $s \le 4,$ y $BC = 17t = 85s.$ Con $s = 4:$ $BC = 340,$ con $BD = 160,$ $CD = 180,$ $AE = 72,$ $AF = 81$ todos enteros positivos, y los lados $200, 225, 340$ forman un triángulo válido puesto que $200 + 225 \gt 340.$

El mayor valor posible de $BC$ es $340.$

Since $\omega$ is tangent to $BC$ at $D,$ the power of $B$ gives $BD^2 = BE \cdot BA$ and the power of $C$ gives $CD^2 = CF \cdot CA.$ The angle bisector gives $\frac{BD}{DC} = \frac{AB}{AC} = \frac{8}{9},$ so $BD = 8t$ and $CD = 9t,$ where $t = CD - BD$ is a positive integer. Then $\begin{aligned} &BE = \frac{64t^2}{200} = \frac{8t^2}{25}, \\ &CF = \frac{81t^2}{225} = \frac{9t^2}{25}, \end{aligned}$ so $AE = 200 - \frac{8t^2}{25}$ and $AF = 225 - \frac{9t^2}{25}.$

For $AE$ and $AF$ to be integers we need $25 \mid t^2,$ that is, $t = 5s.$ Then $AE = 200 - 8s^2 \gt 0$ forces $s \le 4,$ and $BC = 17t = 85s.$ At $s = 4:$ $BC = 340,$ with $BD = 160,$ $CD = 180,$ $AE = 72,$ $AF = 81$ all positive integers, and the sides $200, 225, 340$ form a valid triangle since $200 + 225 \gt 340.$

The greatest possible value of $BC$ is $340.$

2026 AIME II Problema 10

Solución:

El Problema 10 en otros años