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Difference between revisions of "2020 AIME I Problems/Problem 15"
(→‎Solution)
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== Solution ==
 
== 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("<math>B</math>", (-8.914038694762803,3.548005694821766), NE * labelscalefactor);
 +
dot((-0.08068432003432058,3.4366950566657577),dotstyle);
 +
label("<math>C</math>", (-0.021788682572170717,3.594159241597842), NE * labelscalefactor);
 +
dot((-4.538171990791266,4.905585481447388),dotstyle);
 +
label("<math>O</math>", (-4.483298204259513,5.055688222840243), NE * labelscalefactor);
 +
dot((-4.522512329243054,1.9211095752183682),linewidth(4pt) + dotstyle);
 +
label("<math>O'</math>", (-4.467913688667488,2.0403231668032897), NE * labelscalefactor);
 +
dot((-7.790821079477277,5.289342543424063),dotstyle);
 +
label("<math>H</math>", (-7.729430994176854,5.440301112640874), NE * labelscalefactor);
 +
dot((-7.806480741025488,8.273818449653083),linewidth(4pt) + dotstyle);
 +
label("<math>A</math>", (-7.7448155097688804,8.394128106309726), NE * labelscalefactor);
 +
dot((-9.139423209055858,3.980748932978468),linewidth(4pt) + dotstyle);
 +
label("<math>X</math>", (-9.083268366275083,4.101848256134676), NE * labelscalefactor);
 +
dot((-9.752410287411378,3.3859471330788855),linewidth(4pt) + dotstyle);
 +
label("<math>K</math>", (-9.68326447436407,3.5018521480456903), NE * labelscalefactor);
 +
dot((-3.475227037470366,9.476907329794422),linewidth(4pt) + dotstyle);
 +
label("<math>Y</math>", (-3.4063821128177407,9.594120322487697), NE * labelscalefactor);
 +
dot((-7.780888168280388,3.3962917865759925),linewidth(4pt) + dotstyle);
 +
label("<math>L</math>", (-7.714046478584829,3.5172366636377155), NE * labelscalefactor);
 +
dot((-7.776585346090923,2.5762441045932913),linewidth(4pt) + dotstyle);
 +
label("<math>D</math>", (-7.714046478584829,2.7018573372603765), NE * labelscalefactor);
 +
dot((-6.307325123263112,6.728828131386446),linewidth(4pt) + dotstyle);
 +
label("<math>E</math>", (-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 <math>3HX = HY</math> is not insignificant; from here, we set <math>XH = HE = \frac{1}{2} EY = HL = 2</math> by PoP and trivial construction. Now, <math>D</math> is the reflection of <math>A</math> over <math>H</math>. Note <math>AO \perp XY</math>, and therefore by Pythagorean theorem we have <math>AE = XD = \sqrt{5}</math>. Consider <math>HD = 3</math>. We have that <math>\triangle HXD \cong HLK</math>, and therefore we are ready to PoP with respect to <math>(BHC)</math>. Setting <math>BL = x, LC = y</math>, we obtain <math>xy = 10</math> by PoP on <math>(ABC)</math>, and furthermore, we have <math>KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)</math>. Now, we get <math>4 = \sqrt{5}(y - x) - xy</math>, and from <math>xy = 10</math> we take <cmath>\frac{14}{\sqrt{5}} = y - x.</cmath> However, squaring and manipulating with <math>xy = 10</math> yields that <math>(x + y)^2 = \frac{396}{5}</math> and from here, since <math>AL = 5</math> we get the area to be <math>3\sqrt{55} \implies \boxed{058}</math>. ~awang11's sol
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2020|n=I|num-b=14|after=Last Problem}}
 
{{AIME box|year=2020|n=I|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:14, 12 March 2020

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

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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