Question

Sea $\triangle ABC$ un triángulo equilátero de lado $55.$ Los puntos $D,$ $E,$ y $F$ están sobre $\overline{BC},$ $\overline{CA},$ y $\overline{AB},$ respectivamente, con $BD = 7,$ $CE = 30,$ y $AF = 40.$ El punto $P$ dentro de $\triangle ABC$ tiene la propiedad de que $\angle AEP = \angle BFP = \angle CDP.$ Halla $\tan^2(\angle AEP).$

Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE = 30,$ and $AF = 40.$ Point $P$ inside $\triangle ABC$ has the property that $\angle AEP = \angle BFP = \angle CDP.$ Find $\tan^2(\angle AEP).$

Respuesta

Solución:

Coloca $B = (0, 0),$ $C = (55, 0),$ $A = \left(\frac{55}{2}, \frac{55\sqrt{3}}{2}\right),$ de modo que $D = (7, 0),$ $E = (40, 15\sqrt{3}),$ y $F = \left(\frac{15}{2}, \frac{15\sqrt{3}}{2}\right).$ Sea $\theta$ el ángulo común y sean $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ los vectores unitarios en las direcciones $C \to A,$ $A \to B,$ $B \to C,$ es decir, las direcciones desde $E$ hacia $A,$ desde $F$ hacia $B,$ y desde $D$ hacia $C.$ Descomponiendo $P - E$ en su componente a lo largo de $\mathbf{u}_1$ y su componente perpendicular al lado $CA,$ cuya longitud es la distancia $d_1$ de $P$ a la recta $CA,$ la condición angular da $(P - E)\cdot\mathbf{u}_1 = d_1\cot\theta;$ de forma similar $(P - F)\cdot\mathbf{u}_2 = d_2\cot\theta$ y $(P - D)\cdot\mathbf{u}_3 = d_3\cot\theta.$

Ahora suma las tres relaciones. Como $\mathbf{u}_1 + \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}$ (los lados dirigidos de un triángulo se cierran), $P$ desaparece, y por el teorema de Viviani $d_1 + d_2 + d_3$ es igual a la altura $\frac{55\sqrt{3}}{2}.$ Con $\mathbf{u}_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $\mathbf{u}_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right),$ y $\mathbf{u}_3 = (1, 0),$ obtenemos $E \cdot \mathbf{u}_1 = \frac{5}{2},$ $F \cdot \mathbf{u}_2 = -15,$ y $D \cdot \mathbf{u}_3 = 7,$ así que $\begin{aligned} \frac{55\sqrt{3}}{2}\,\cot\theta &= -\left(\frac{5}{2} - 15 + 7\right) \\ &= \frac{11}{2}. \end{aligned}$

Por lo tanto $\cot\theta = \frac{11}{55\sqrt{3}} = \frac{1}{5\sqrt{3}},$ así que $\tan^2(\angle AEP) = \left(5\sqrt{3}\right)^2 = 75.$

Place $B = (0, 0),$ $C = (55, 0),$ $A = \left(\frac{55}{2}, \frac{55\sqrt{3}}{2}\right),$ so that $D = (7, 0),$ $E = (40, 15\sqrt{3}),$ and $F = \left(\frac{15}{2}, \frac{15\sqrt{3}}{2}\right).$ Let $\theta$ be the common angle and let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ be the unit vectors in the directions $C \to A,$ $A \to B,$ $B \to C$ — the directions from $E$ toward $A,$ from $F$ toward $B,$ and from $D$ toward $C.$ Splitting $P - E$ into its component along $\mathbf{u}_1$ and its component perpendicular to side $CA,$ whose length is the distance $d_1$ from $P$ to line $CA,$ the angle condition gives $(P - E)\cdot\mathbf{u}_1 = d_1\cot\theta;$ similarly $(P - F)\cdot\mathbf{u}_2 = d_2\cot\theta$ and $(P - D)\cdot\mathbf{u}_3 = d_3\cot\theta.$

Now add all three relations. Since $\mathbf{u}_1 + \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}$ (the directed sides of a triangle close up), $P$ drops out, and by Viviani's theorem $d_1 + d_2 + d_3$ equals the height $\frac{55\sqrt{3}}{2}.$ With $\mathbf{u}_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $\mathbf{u}_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right),$ and $\mathbf{u}_3 = (1, 0),$ we get $E \cdot \mathbf{u}_1 = \frac{5}{2},$ $F \cdot \mathbf{u}_2 = -15,$ and $D \cdot \mathbf{u}_3 = 7,$ so $\begin{aligned} \frac{55\sqrt{3}}{2}\,\cot\theta &= -\left(\frac{5}{2} - 15 + 7\right) \\ &= \frac{11}{2}. \end{aligned}$

Hence $\cot\theta = \frac{11}{55\sqrt{3}} = \frac{1}{5\sqrt{3}},$ so $\tan^2(\angle AEP) = \left(5\sqrt{3}\right)^2 = 75.$

2023 AIME I Problema 12

Solución:

El Problema 12 en otros años