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2020 AIME I Problems/Problem 13

Revision as of 16:57, 12 March 2020 by Vvluo (talk | contribs) (→‎Solution)

Problem

Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$


[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 = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq); draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227)--(-3.319253031309944,4.210570466954303)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-6.837129089839387,8.163360372429347)--(-7.3192122908832715,4.192517163831042), linewidth(2) + wrwrwr); draw((-7.3192122908832715,4.192517163831042)--(-2.319263216416622,4.2150837927351175), linewidth(2) + wrwrwr); draw((-2.319263216416622,4.2150837927351175)--(-6.837129089839387,8.163360372429347), linewidth(2) + wrwrwr); draw((xmin, -2.6100704119306224*xmin-9.68202796751058)--(xmax, -2.6100704119306224*xmax-9.68202796751058), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.3831314264278095*xmin + 8.511194202815297)--(xmax, 0.3831314264278095*xmax + 8.511194202815297), linewidth(2) + wrwrwr); /* line */ draw(circle((-6.8268938290378,5.895596632024835), 2.267786838055365), linewidth(2) + wrwrwr); draw(circle((-4.33118398380513,6.851781504978754), 2.828427124746193), linewidth(2) + wrwrwr); draw((xmin, 0.004513371749987873*xmin + 4.225551489816879)--(xmax, 0.004513371749987873*xmax + 4.225551489816879), linewidth(2) + wrwrwr); /* line */ draw((-7.3192122908832715,4.192517163831042)--(-4.33118398380513,6.851781504978754), linewidth(2) + wrwrwr); draw((-6.8268938290378,5.895596632024835)--(-2.319263216416622,4.2150837927351175), linewidth(2) + wrwrwr); draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227), linewidth(2) + wrwrwr); draw((xmin, 0.004513371749987873*xmin + 8.19421887771445)--(xmax, 0.004513371749987873*xmax + 8.19421887771445), linewidth(2) + wrwrwr); /* line */ draw((-3.837159645159393,8.176900349771794)--(-8.31920210577661,4.188003838050227), linewidth(2) + wrwrwr); draw((-3.837159645159393,8.176900349771794)--(-5.3192326610966125,4.2015438153926725), linewidth(2) + wrwrwr); draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835), linewidth(2) + rvwvcq); draw((-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754), linewidth(2) + rvwvcq); draw((-4.33118398380513,6.851781504978754)--(-6.837129089839387,8.163360372429347), linewidth(2) + rvwvcq); draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227), linewidth(2) + rvwvcq); draw((-8.31920210577661,4.188003838050227)--(-3.319253031309944,4.210570466954303), linewidth(2) + rvwvcq); draw((-3.319253031309944,4.210570466954303)--(-6.837129089839387,8.163360372429347), linewidth(2) + rvwvcq); /* dots and labels */ dot((-6.837129089839387,8.163360372429347),dotstyle); label("$A$", (-6.8002301023571095,8.267690318323321), NE * labelscalefactor); dot((-7.3192122908832715,4.192517163831042),dotstyle); label("$B$", (-7.2808283997985,4.29753046989445), NE * labelscalefactor); dot((-2.319263216416622,4.2150837927351175),linewidth(4pt) + dotstyle); label("$C$", (-2.276337432963145,4.29753046989445), NE * labelscalefactor); dot((-5.3192326610966125,4.2015438153926725),linewidth(4pt) + dotstyle); label("$D$", (-5.274852897434433,4.287082680819637), NE * labelscalefactor); dot((-6.8268938290378,5.895596632024835),linewidth(4pt) + dotstyle); label("$F$", (-6.789782313282296,5.979624510939313), NE * labelscalefactor); dot((-4.33118398380513,6.851781504978754),linewidth(4pt) + dotstyle); label("$E$", (-4.292760724402025,6.93037331674728), NE * labelscalefactor); dot((-8.31920210577661,4.188003838050227),linewidth(4pt) + dotstyle); label("$G$", (-8.273368361905721,4.276634891744824), NE * labelscalefactor); dot((-3.319253031309944,4.210570466954303),linewidth(4pt) + dotstyle); label("$H$", (-3.2793251841451787,4.29753046989445), NE * labelscalefactor); dot((-3.837159645159393,8.176900349771794),linewidth(4pt) + dotstyle); label("$I$", (-3.7912668488110084,8.257242529248508), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Solution 2(coordinate bash based)

Let $\overline{BC}$ lie on the x-axis and $B$ be the origin. $C$ is $(5,0)$. Use Heron's formula to compute the area of triangle $ABC$. We have $s=\frac{15}{2}$. and $[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}$. We now find the altitude, which is $\frac{\frac{15\sqrt{7}}{2}}{5}=3\sqrt{7}{2}$, which is the y-coordinate of $A$. We now find the x-coordinate of $A$, which satisfies $x^2 + (3\sqrt{7}{2})^2=16$, which gives $x=\frac{1}{2}$ since the triangle is acute. Now using the Angle Bisector Theorem, we have $\frac{4}{6}=\frac{BD}{CD}$ and $BD+CD=5$ to get $BD=2$. The coordinates of D are $(2,0)$. Since we want the area of triangle $AEF$, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is $(\frac{5}{4}, \frac{3\sqrt{7}}{4})$ and the slope of AD is $-\sqrt{7}$. The slope of the perpendicular bisector is $\frac{1}{\sqrt{7}}$. The equation is(in point slope form) $y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})$. The slope of AB, or in trig words, the tangent of $\angle ABC$ is $3\sqrt{7}$. Finding $\sin{\angle ABC}=\frac{\frac{3\sqrt{7}}{2}}{4}=\frac{3\sqrt{7}}{8}$ and $\cos{\angle ABC}=\frac{\frac{1}{2}}{4}=\frac{1}{8}$. Plugging this in to half angle tangent, it gives $\frac{\frac{3\sqrt{7}}{8}}{1+\frac{1}{8}}=\frac{\sqrt{7}}{3}$ as the slope of the angle bisector, since it passes through $B$, the equation is $y=\frac{\sqrt{7}}{3}x$. Similarly, the equation for the angle bisector of $C$ will be $y=-\frac{1}{\sqrt{7}}(x-5).

Points are defined as shown. It is pretty easy to show that $\triangle AFE \sim \triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\triangle AFE$, as the altitude of $AGH$. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that $AD/2 = \frac{\sqrt{18}}{2}$, the altitude of $\triangle AFE$. Similarly, the altitude of $\triangle AGH$ is the altitude of $\triangle ABC$, or $\frac{12}{\sqrt{7}}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\triangle ABC$ is $\frac{15\sqrt{7}}{14} \implies \boxed{036}$, by similarity. ~awang11

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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