Question

Sea $\triangle ABC$ con lados $AB = 13,$ $BC = 14,$ y $CA = 15.$ El triángulo $\triangle A'B'C'$ se obtiene rotando $\triangle ABC$ alrededor de su circuncentro de modo que $\overline{A'C'}$ sea perpendicular a $\overline{BC},$ con $A'$ y $B$ no del mismo lado de la recta $B'C'.$ Halle el entero más cercano al área del hexágono $AA'CC'BB'.$

Let $\triangle ABC$ have side lengths $AB = 13,$ $BC = 14,$ and $CA = 15.$ Triangle $\triangle A'B'C'$ is obtained by rotating $\triangle ABC$ about its circumcenter so that $\overline{A'C'}$ is perpendicular to $\overline{BC},$ with $A'$ and $B$ not on the same side of line $B'C'.$ Find the integer closest to the area of hexagon $AA'CC'BB'.$

Respuesta

Solución:

Coloque $B = (0,0),$ $C = (14,0),$ $A = (5,12).$ El circuncentro está en $x = 7,$ e igualando las distancias a $B$ y $A$ se obtiene $O = \left(7, \frac{33}{8}\right).$ La dirección de $\overline{AC}$ es $C - A = (9, -12),$ paralela a $(3, -4).$ Una rotación de ángulo $\varphi$ hace $\overline{A'C'}$ vertical exactamente cuando envía $(3,-4)$ a $(0, \pm 5),$ así que $(\cos\varphi, \sin\varphi) = \left(\frac{4}{5}, -\frac{3}{5}\right)$ o $\left(-\frac{4}{5}, \frac{3}{5}\right).$ Rotando cada vértice alrededor de $O$ y verificando la recta $B'C'$ se muestra que $A'$ y $B$ están en lados opuestos solo para $\cos\varphi = \frac{4}{5},$ $\sin\varphi = -\frac{3}{5}.$

Con esta rotación, $P' = O + R(P - O)$ da $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $B' = \left(-\tfrac{43}{40}, \tfrac{201}{40}\right),$ $C' = \left(\tfrac{81}{8}, -\tfrac{27}{8}\right).$ Por ejemplo, $A - O = \left(-2, \tfrac{63}{8}\right)$ se rota a $\left(\tfrac{25}{8}, \tfrac{15}{2}\right),$ dando $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right).$

El hexágono $A A' C C' B B'$ es simple con estos vértices en orden, así que la fórmula de la lazada sobre $(5,12),$ $\left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $(14,0),$ $\left(\tfrac{81}{8}, -\tfrac{27}{8}\right),$ $(0,0),$ $\left(-\tfrac{43}{40}, \tfrac{201}{40}\right)$ da área $\frac{1557}{10} = 155.7.$ El entero más cercano es $156.$

Place $B = (0,0),$ $C = (14,0),$ $A = (5,12).$ The circumcenter lies on $x = 7,$ and equating distances to $B$ and $A$ gives $O = \left(7, \frac{33}{8}\right).$ The direction of $\overline{AC}$ is $C - A = (9, -12),$ parallel to $(3, -4).$ A rotation through $\varphi$ makes $\overline{A'C'}$ vertical exactly when it sends $(3,-4)$ to $(0, \pm 5),$ so $(\cos\varphi, \sin\varphi) = \left(\frac{4}{5}, -\frac{3}{5}\right)$ or $\left(-\frac{4}{5}, \frac{3}{5}\right).$ Rotating each vertex about $O$ and checking the line $B'C'$ shows that $A'$ and $B$ are on opposite sides only for $\cos\varphi = \frac{4}{5},$ $\sin\varphi = -\frac{3}{5}.$

With this rotation, $P' = O + R(P - O)$ gives $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $B' = \left(-\tfrac{43}{40}, \tfrac{201}{40}\right),$ $C' = \left(\tfrac{81}{8}, -\tfrac{27}{8}\right).$ For example, $A - O = \left(-2, \tfrac{63}{8}\right)$ rotates to $\left(\tfrac{25}{8}, \tfrac{15}{2}\right),$ giving $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right).$

The hexagon $A A' C C' B B'$ is simple with these vertices in order, so the shoelace formula on $(5,12),$ $\left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $(14,0),$ $\left(\tfrac{81}{8}, -\tfrac{27}{8}\right),$ $(0,0),$ $\left(-\tfrac{43}{40}, \tfrac{201}{40}\right)$ gives area $\frac{1557}{10} = 155.7.$ The closest integer is $156.$

2026 AIME I Problema 10

Solución:

El Problema 10 en otros años