Question

Sea $\triangle ABC$ un triángulo acutángulo escaleno con circunferencia circunscrita $\omega.$ Las tangentes a $\omega$ en $B$ y $C$ se cortan en $T.$ Sean $X$ y $Y$ las proyecciones de $T$ sobre las rectas $AB$ y $AC,$ respectivamente. Suponga que $BT = CT = 16,$ $BC = 22,$ y $TX^2 + TY^2 + XY^2 = 1143.$ Halle $XY^2.$

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega.$ The tangents to $\omega$ at $B$ and $C$ intersect at $T.$ Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC,$ respectively. Suppose $BT = CT = 16,$ $BC = 22,$ and $TX^2 + TY^2 + XY^2 = 1143.$ Find $XY^2.$

Respuesta

Solución:

Por el ángulo tangente-cuerda, $\angle TBC = A,$ así que $\angle ABT = B + A = 180^\circ - C$ y $TX = TB \sin\angle ABT = 16 \sin C;$ de manera similar $TY = 16 \sin B.$ Además $\angle AXT = \angle AYT = 90^\circ,$ así que $A, X, T, Y$ están sobre una circunferencia de diámetro $AT,$ de donde $XY = AT \sin A.$ Usando la ley de senos $\scriptsize\left(\sin A = \frac{11}{R},\ \sin B = \frac{AC}{2R},\ \sin C = \frac{AB}{2R}\right),$ la condición dada se convierte en $\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}$

Coloque $B = (-11, 0)$ y $C = (11, 0).$ Como $TB = 16$ y $T$ está en la mediatriz de $BC,$ obtenemos $T = (0, -\sqrt{135}).$ El circuncentro es $O = (0, k)$ con $OB \perp BT,$ lo que da $121 - k\sqrt{135} = 0,$ así que $k = \frac{121}{\sqrt{135}}$ y $R^2 = 121 + k^2 = \frac{30976}{135}.$ Para $A = (x, y)$ sobre $\omega,$ desarrollando $x^2 + (y - k)^2 = R^2$ se obtiene $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.$ Por lo tanto $\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}$ $\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}$

Sustituyendo, $64(AB^2 + AC^2) + 121\,AT^2$ $= \frac{92928}{\sqrt{135}}\,y + 61952$ $= 1143 \cdot \frac{30976}{135}$ produce $y = \frac{291}{\sqrt{135}}.$ Entonces $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},$ y $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$

By the tangent-chord angle, $\angle TBC = A,$ so $\angle ABT = B + A = 180^\circ - C$ and $TX = TB \sin\angle ABT = 16 \sin C;$ similarly $TY = 16 \sin B.$ Also $\angle AXT = \angle AYT = 90^\circ,$ so $A, X, T, Y$ lie on a circle with diameter $AT,$ whence $XY = AT \sin A.$ Using the law of sines $\scriptsize\left(\sin A = \frac{11}{R},\ \sin B = \frac{AC}{2R},\ \sin C = \frac{AB}{2R}\right),$ the given condition becomes $\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}$

Place $B = (-11, 0)$ and $C = (11, 0).$ Since $TB = 16$ and $T$ lies on the perpendicular bisector of $BC,$ we get $T = (0, -\sqrt{135}).$ The circumcenter is $O = (0, k)$ with $OB \perp BT,$ which gives $121 - k\sqrt{135} = 0,$ so $k = \frac{121}{\sqrt{135}}$ and $R^2 = 121 + k^2 = \frac{30976}{135}.$ For $A = (x, y)$ on $\omega,$ expanding $x^2 + (y - k)^2 = R^2$ gives $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.$ Therefore $\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}$ $\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}$

Substituting, $64(AB^2 + AC^2) + 121\,AT^2$ $= \frac{92928}{\sqrt{135}}\,y + 61952$ $= 1143 \cdot \frac{30976}{135}$ yields $y = \frac{291}{\sqrt{135}}.$ Then $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},$ and $\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$

2020 AIME II Problema 15

Solución:

El Problema 15 en otros años