Difference between revisions of "2020 AIME I Problems/Problem 15"

(Problem)
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
Note: Please do not post problems here until after the AIME.
 
  
 
== Problem ==
 
== Problem ==
Let <math>\triangle ABC</math> be an acute triangle with circumcircle <math>\omega,</math> and let <math>H</math> be the intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> intersects <math>\omega</math> at points <math>X</math> and <math>Y</math> with <math>HA=3,HX=2,</math> and <math>HY=6.</math> The area of <math>\triangle ABC</math> can be written as <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math>
+
Let <math>\triangle ABC</math> be an acute triangle with circumcircle <math>\omega,</math> and let <math>H</math> be the intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> intersects <math>\omega</math> at points <math>X</math> and <math>Y</math> with <math>HA=3,HX=2,</math> and <math>HY=6.</math> The area of <math>\triangle ABC</math> can be written in the form <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math>
  
== Solution ==
+
== Solution 1==
 
The following is a power of a point solution to this menace of a problem:
 
The following is a power of a point solution to this menace of a problem:
 
<asy>
 
<asy>
Line 53: Line 52:
  
 
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
 
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
 +
 +
==Solution 1a==
 +
As in the diagram, let ray <math>AH</math> extended hits BC at L and the circumcircle at say <math>P</math>. By power of the point at H, we have <math>HX \cdot HY = AH \cdot HP</math>. The three values we are given tells us that <math>HP=\frac{2\cdot 6}{3}=4</math>. L is the midpoint of <math>HP</math>(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so <math>HL=LP=2</math>.
 +
 +
As in the diagram provided, let K be the intersection of <math>BC</math> and <math>XY</math>. By power of a point on the circumcircle of triangle <math>HBC</math>, <math>KH^{2}=KB \cdot KC</math>. By power of a point on the circumcircle of triangle <math>ABC</math>, <math>KB \cdot KC=KX \cdot KY</math>, thus <math>KH^{2}=(KH-2)(KH+6)</math>. Solving gives <math>4KH=12</math> or <math>KH=3</math>.
 +
 +
By the Pythagorean Theorem on triangle <math>HKL</math>, <math>KL=\sqrt{5}</math>. Now continue with solution 1.
  
 
== Solution 2 ==
 
== 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 <math>H'</math>, the image of the reflection of <math>H</math> over line <math>BC</math>, lies on circle <math>O</math>.  This is because <math>\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC</math>.  This is a well known lemma.  The result of this observation is that circle <math>O'</math>, the circumcircle of <math>\triangle BHC</math> is the image of circle <math>O</math> over line <math>BC</math>, which in turn implies that <math>\overline{AH} = \overline{OO'}</math> and thus <math>AHO'O</math> is a parallelogram.  That <math>AHO'O</math> is a parallelogram implies that <math>AO</math> is perpendicular to <math>\overline{XY}</math>, and thus divides segment <math>\overline{XY}</math> in two equal pieces, <math>\overline{XD}</math> and <math>\overline{DY}</math>, of length <math>4</math>.
 +
 +
 +
Using Power of a Point,
 +
<cmath>\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4</cmath>
 +
This means that <math>\overline{HL} = \frac12 \cdot 4 = 2</math> and <math>\overline{AL} = 2 + 3 = 5</math>, where <math>L</math> is the foot of the altitude from <math>A</math> onto <math>BC</math>.  All that remains to be found is the length of segment <math>\overline{BC}</math>.
 +
 +
Looking at right triangle <math>\triangle AHD</math>, we find that
 +
<cmath>\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}</cmath>
 +
Looking at right triangle <math>\triangle ODY</math>, we get the equation
 +
<cmath>\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2</cmath>
 +
Plugging in known values, and letting <math>R</math> be the radius of the circle, we find that
 +
<cmath>R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}</cmath>
 +
 +
Recall that <math>AHO'O</math> is a parallelogram, so <math>\overline{AH} = \overline{OO'} = 3</math>.  So, <math>\overline{OM} = \frac32</math>, where <math>M</math> is the midpoint of <math>\overline{BC}</math>.  This means that
 +
<cmath>\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}</cmath>
 +
 +
Thus, the area of triangle <math>\triangle ABC</math> is
 +
<cmath>\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}</cmath>
 +
The answer is <math>3 + 55 = \boxed{058}</math>.
 +
 +
== Video Solution ==
 +
https://www.youtube.com/watch?v=L7B20E95s4M
  
 
==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 15:28, 31 August 2020

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 in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$

Solution 1

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 1a

As in the diagram, let ray $AH$ extended hits BC at L and the circumcircle at say $P$. By power of the point at H, we have $HX \cdot HY = AH \cdot HP$. The three values we are given tells us that $HP=\frac{2\cdot 6}{3}=4$. L is the midpoint of $HP$(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so $HL=LP=2$.

As in the diagram provided, let K be the intersection of $BC$ and $XY$. By power of a point on the circumcircle of triangle $HBC$, $KH^{2}=KB \cdot KC$. By power of a point on the circumcircle of triangle $ABC$, $KB \cdot KC=KX \cdot KY$, thus $KH^{2}=(KH-2)(KH+6)$. Solving gives $4KH=12$ or $KH=3$.

By the Pythagorean Theorem on triangle $HKL$, $KL=\sqrt{5}$. Now continue with solution 1.

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 \[\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2\] 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 \[\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}\] The answer is $3 + 55 = \boxed{058}$.

Video Solution

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

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png