2020 AIME I Problems/Problem 13

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 1

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

Solution 2(coord bash + basic geometry)

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 $\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}$ , which is the y-coordinate of $A$ . We now find the x-coordinate of $A$ , which satisfies $x^2 + (\frac{3\sqrt{7}}{2})^{2}=16$ , 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 $y=-\frac{1}{\sqrt{7}}(x-5)$ . For $E$ use the B-angle bisector and the perpendicular bisector of AD equations to intersect at $(3,\sqrt{7})$ . For $F$ use the C-angle bisector and the perpendicular bisector of AD equations to intersect at $(\frac{1}{2}, \frac{9}{2\sqrt{7}})$ . The area of AEF is equal to $\frac{EF \cdot \frac{AD}{2}}{2}$ since AD is the altitude of that triangle with EF as the base, with $\frac{AD}{2}$ being the height. $EF=\frac{5\sqrt{2}}{\sqrt{7}}$ and $AD=3\sqrt{2}$ , so $[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}$ which gives $\boxed{036}$ . NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo

Solution 3 (Coordinate Bash + Trig)

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); [/asy]$

Let $B=(0,0)$ and $BC$ be the line $y=0$ . We compute that $\cos{\angle{ABC}}=\frac{1}{8}$ , so $\tan{\angle{ABC}}=3\sqrt{7}$ . Thus, $A$ lies on the line $y=3x\sqrt{7}$ . The length of $AB$ at a point $x$ is $8x$ , so $x=\frac{1}{2}$ .

We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$ , $B=(0,0)$ and $C=(5,0)$ . We also have $D=(2,0)$ by the angle-bisector theorem and $M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ by taking the midpoint. We have that because $\cos{\angle{ABC}}=\frac{1}{8}$ , $\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}$ by half angle formula.

We also compute $\cos{\angle{ACB}}=\frac{3}{4}$ , so $\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}$ .

Now, $AD$ has slope $-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}$ , so it's perpendicular bisector has slope $\frac{\sqrt{7}}{7}$ and goes through $\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ .

We find that this line has equation $y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}$ .

As $\cos{\angle{CBI}}=\frac{3}{4}$ , we have that line $BI$ has form $y=\frac{\sqrt{7}}{3}x$ . Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\left(3, \sqrt{7}\right)$

We also have that because $\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}$ , $CI$ has form $y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$ .

Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}$ .

Solving this, we get $x=\frac{1}{2}$ and so $F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)$ .

We now compute $EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}$ . We also have $MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}$ .

As ${MA}\perp{EF}$ , we have $[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}$ .

The desired answer is $15+7+14=\boxed{036}$ ~Imayormaynotknowcalculus

Solution 4 (Barycentric Coordinates)

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); [/asy]$

As usual, we will use homogenized barycentric coordinates.

We have that $AD$ will have form $3z=2y$ . Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$ . Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$ , we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ . It remains to determine the equation of the line formed by the perpendicular bisector of $AD$ .

This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$ . We then have that the displacement vector $\overrightarrow{AD}=\left(-1, \frac{3}{5}, \frac{2}{5}\right)$ and that the displacement vector $\overrightarrow{TM}$ has form $\left(x-\frac{1}{2},y-\frac{3}{10},z-\frac{1}{5}\right)$ . Now, by EFFT, we have $5^2\left(\frac{3}{5}\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(y-\frac{3}{10}\right)\right)+6^2\left(-1\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(x-\frac{1}{2}\right)\right)+4^2\left(-1\times\left(y-\frac{3}{10}\right)+\frac{3}{5}\times\left(x-\frac{1}{2}\right)\right)=0$ . This equates to $8x-2y-7z=2$ .

Now, intersecting this with $BE$ , we have $5z=4x$ , $8x-2y-7z=2$ , and $x+y+z=1$ . This yields $x=\frac{2}{3}$ , $y=-\frac{1}{5}$ , and $z=\frac{8}{15}$ , or $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$ .

Similarly, intersecting this with $CF$ , we have $5y=6x$ , $8x-2y-7z=2$ , and $x+y+z=1$ . Solving this, we obtain $x=\frac{3}{7}$ , $y=\frac{18}{35}$ , and $z=\frac{2}{35}$ , or $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$ .

We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\overrightarrow{FE}=\left(\frac{5}{21},-\frac{5}{7},\frac{10}{21}\right)$ . We then have $FE^2=-25\left(-\frac{5}{7}\cdot\frac{10}{21}\right)-36\left(\frac{5}{21}\cdot\frac{10}{21}\right)-16\left(\frac{5}{21}\cdot-\frac{5}{7}\right)=\frac{50}{7}$ , thus $FE=\frac{5\sqrt{14}}{7}$ .

Our second displacement vector is $\overrightarrow{AM}=\left(-\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ . As a result, $AM^2=-25\left(\frac{3}{10}\cdot\frac{1}{5}\right)-36\left(-\frac{1}{2}\cdot\frac{1}{5}\right)-16\left(-\frac{1}{2}\cdot\frac{3}{10}\right)=\frac{9}{2}$ , so $AM=\frac{3\sqrt{2}}{2}$ .

As ${AM}\perp{EF}$ , the desired area is $\frac{\frac{5\sqrt{14}}{7}\times\frac{3\sqrt{2}}{2}}{2}={\frac{15\sqrt{7}}{14}}\implies{m+n+p=\boxed{036}}$ . ~Imayormaynotknowcalculus

Remark: The area of $\triangle{AEF}$ can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.

Solution 5 (geometry+trig)

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); [/asy]$

To get the area of $\triangle AEF$ , we try to find $AM$ and $\angle EAF$ .

Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$ . By applying Stewart's Theorem, we can get that $AD=3\sqrt{2}$ . Therefore $AM=\frac{3\sqrt{2}}{2}$ .

Since $EF$ is the perpendicular bisector of $AD$ , we know that $AE = DE$ . Since $BE$ is the angle bisector of $\angle BAC$ , we know that $\angle ABE = \angle DBE$ . By applying the Law of Sines to $\triangle ABE$ and $\triangle DBE$ , we know that $\sin \angle BAE = \sin \angle BDE$ . Since $BD$ is not equal to $AB$ and therefor these two triangles are not congruent, we know that $\angle BAE$ and $\angle BDE$ are complement. Then we know that $\angle ABD$ and $\angle AED$ are also complement. Given that $AE=DE$ , we can get that $\angle DAE$ is half of $\angle ABC$ . Similarly, we have $\angle DAF$ is half of $\angle ACB$ .

By applying the Law of Cosines, we get $\cos \angle ABC = \frac{1}{8}$ , and then $\sin \angle ABC = \frac{3\sqrt{7}}{8}$ . Similarly, we can get $\cos \angle ACB = \frac{3}{4}$ and $\sin \angle ACB = \frac{\sqrt{7}}{4}$ . Based on some trig identities, we can compute that $\tan \angle DAE = \frac{\sin \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}$ , and $\tan \angle DAF = \frac{\sqrt{7}}{7}$ .

Finally, the area of $\triangle AEF$ equals $\frac{1}{2}AM^2(\tan \angle DAE + \tan \angle DAF)=\frac{15\sqrt{7}}{14}$ . Therefore, the final answer is $15+7+14=\boxed{036}$ . ~xamydad

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