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2020 AIME I Problems/Problem 15

Revision as of 21:39, 13 March 2020 by Huangyi 99 (talk | contribs) (→‎Solution 2)

Problem

Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written as $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$

Solution

The following is a power of a point solution to this menace of a problem: [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.821705655137235, xmax = 10.870448356581754, ymin = -3.0673360097491003, ymax = 10.363346102088961; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((xmin, 0.0052470390246834855*xmin + 3.437118410441658)--(xmax, 0.0052470390246834855*xmax + 3.437118410441658), linewidth(2) + wrwrwr); /* line */ draw(circle((-4.538171990791266,4.905585481447388), 4.693275552848494), linewidth(2) + wrwrwr); draw(circle((-4.522512329243054,1.9211095752183682), 4.693275552848494), linewidth(2) + wrwrwr); draw((xmin, -190.58367877496823*xmin-1479.5139994609244)--(xmax, -190.58367877496823*xmax-1479.5139994609244), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.9703333412757664*xmin + 12.849035992754926)--(xmax, 0.9703333412757664*xmax + 12.849035992754926), linewidth(2) + wrwrwr); /* line */ draw(circle((-7.790821079477277,5.289342543424063), 1.8930768158550504), linewidth(2) + wrwrwr); draw((xmin, -1.0305736775830343*xmin-5.438100054565965)--(xmax, -1.0305736775830343*xmax-5.438100054565965), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.0305736775830343*xmin + 0.22866488339331612)--(xmax, -1.0305736775830343*xmax + 0.22866488339331612), linewidth(2) + wrwrwr); /* line */ /* dots and labels */ dot((-8.98,3.39),dotstyle); label("$B$", (-8.914038694762803,3.548005694821766), NE * labelscalefactor); dot((-0.08068432003432058,3.4366950566657577),dotstyle); label("$C$", (-0.021788682572170717,3.594159241597842), NE * labelscalefactor); dot((-4.538171990791266,4.905585481447388),dotstyle); label("$O$", (-4.483298204259513,5.055688222840243), NE * labelscalefactor); dot((-4.522512329243054,1.9211095752183682),linewidth(4pt) + dotstyle); label("$O'$", (-4.467913688667488,2.0403231668032897), NE * labelscalefactor); dot((-7.790821079477277,5.289342543424063),dotstyle); label("$H$", (-7.729430994176854,5.440301112640874), NE * labelscalefactor); dot((-7.806480741025488,8.273818449653083),linewidth(4pt) + dotstyle); label("$A$", (-7.7448155097688804,8.394128106309726), NE * labelscalefactor); dot((-9.139423209055858,3.980748932978468),linewidth(4pt) + dotstyle); label("$X$", (-9.083268366275083,4.101848256134676), NE * labelscalefactor); dot((-9.752410287411378,3.3859471330788855),linewidth(4pt) + dotstyle); label("$K$", (-9.68326447436407,3.5018521480456903), NE * labelscalefactor); dot((-3.475227037470366,9.476907329794422),linewidth(4pt) + dotstyle); label("$Y$", (-3.4063821128177407,9.594120322487697), NE * labelscalefactor); dot((-7.780888168280388,3.3962917865759925),linewidth(4pt) + dotstyle); label("$L$", (-7.714046478584829,3.5172366636377155), NE * labelscalefactor); dot((-7.776585346090923,2.5762441045932913),linewidth(4pt) + dotstyle); label("$D$", (-7.714046478584829,2.7018573372603765), NE * labelscalefactor); dot((-6.307325123263112,6.728828131386446),linewidth(4pt) + dotstyle); label("$E$", (-6.252517497342424,6.855676547107199), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$. Note $AO \perp XY$, and therefore by Pythagorean theorem we have $AE = XD = \sqrt{5}$. Consider $HD = 3$. We have that $\triangle HXD \cong HLK$, and therefore we are ready to PoP with respect to $(BHC)$. Setting $BL = x, LC = y$, we obtain $xy = 10$ by PoP on $(ABC)$, and furthermore, we have $KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)$. Now, we get $4 = \sqrt{5}(y - x) - xy$, and from $xy = 10$ we take \[\frac{14}{\sqrt{5}} = y - x.\] However, squaring and manipulating with $xy = 10$ yields that $(x + y)^2 = \frac{396}{5}$ and from here, since $AL = 5$ we get the area to be $3\sqrt{55} \implies \boxed{058}$. ~awang11's sol

Solution 2

[asy] size(10cm); pair A, B, C, D, H, K, O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label("$O$", O, ENE); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, E); label("$H$", H, E); label("$H'$", K, NE); label("$X$", X, W); label("$Y$", Y, NE); label("$O'$", P, E); label("$M$", M, NE); label("$L$", L, NE); label("$D$", D, NNE); label("$2$", X -- H, NW); label("$3$", H -- A, SW); label("$6$", H -- Y, NW); label("$R$", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy] Diagram not to scale.


We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\triangle BHC$ is the image of circle $O$ over line $BC$, which in turn implies that $\overline{AH} = \overline{OO'}$ and thus $AHO'O$ is a parallelogram. That $AHO'O$ is a parallelogram implies that $AO$ is perpendicular to $\overline{XY}$, and thus divides segment $\overline{XY}$ in two equal pieces, $\overline{XD}$ and $\overline{DY}$, of length $4$.


Using Power of a Point, \[\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4\] This means that $\overline{HL} = \frac12 \cdot 4 = 2$ and $\overline{AL} = 2 + 3 = 5$, where $L$ is the foot of the altitude from $A$ onto $BC$. All that remains to be found is the length of segment $\overline{BC}$.

Looking at right triangle $\triangle AHD$, we find that \[\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\] Looking at right triangle $\triangle ODY$, we get the equation \[\overline{OY}^2 - \overline{HY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2\] Plugging in known values, and letting $R$ be the radius of the circle, we find that \[R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}\]

Recall that $AHO'O$ is a parallelogram, so $\overline{AH} = \overline{OO'} = 3$. So, $\overline{OM} = \frac32$, where $M$ is the midpoint of $\overline{BC}$. This means that \[\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}\]

Thus, the area of triangle $\triangle ABC$ is \[\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = \boxed{3\sqrt{55}}\] The answer is $3 + 55 = \boxed{058}$.

Video Solution

https://www.youtube.com/watch?v=L7B20E95s4M

See Also

2020 AIME I (Problems • Answer Key • Resources)
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