2020 AIME I Problems/Problem 15

Note: Please do not post problems here until after the AIME.

Problem

Solution

The following is a simple 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

2020 AIME I (Problems • Answer Key • Resources)
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2020 AIME I Problems/Problem 15

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