2020 AIME I Problems/Problem 13

1 Problem
2 Diagram
3 Solution 1
4 Solution 2
5 Solution 3(coord bash + basic geometry)
6 Solution 4 (Coordinate Bash + Trig)
7 Solution 5 (Barycentric Coordinates)
8 Solution 6 (geometry+trig)
9 Solution 7
10 Solution 8 -Trigonometry(only)
11 Solution 9 (Official MAA 1)
12 Solution 10 (Official MAA 2)
13 Solution 11 Bash for life
14 Solution 12 (Simple geometry)
15 Video Solution
16 See Also

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 $m+n+p.$

Diagram

$[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(0.8235294117647059,0.0396078431372549,0.07529411764705882); 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 $\frac{3\sqrt{7}}{2}$ . 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

Let $M_A$ , $M_B$ , $M_C$ be the midpoints of arcs $BC$ , $CA$ , $AB$ . By Fact 5, we know that $M_AB = M_AC = M_AI$ , and so by Ptolmey's theorem, we deduce that $AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.$ In particular, we have $AI = IM_A$ . $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot("$A$", A, dir(120)); dot("$B$", B, dir(220)); dot("$C$", C, dir(320)); dot("$D$", D, dir(230)); dot("$E$", E, dir(330)); dot("$F$", F, dir(250)); dot("$I$", I, dir(80)); dot("$M_A$", MA, dir(270)); dot("$M_B$", MB, dir(60)); dot("$M_C$", MC, dir(150)); [/asy]$ Now the key claim is that:

Claim: $\triangle DEF$ and $\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$ .

Proof. First, we show that $D$ is the midpoint of $M_AI$ . Indeed, we have $\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1$ by Ratio lemma and Law of Sines.

Now observe that:

$\overline{M_BM_C}$ is the perpendicular bisector of $\overline{AI}$ ,
$\overline{EF}$ is the perpendicular bisector of $\overline{AD}$ , and
$ID = AI/2$ .

Combining these facts gives that $\overline{EF}$ is a midline in $\triangle IM_BM_C$ , which proves the claim. $\blacksquare$

To finish, we compute $[M_AM_BM_C]$ , noting that $[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]$ .

By Heron's, we can calculate the circumradius $R = 8/\sqrt{7}$ , and by Law of Cosines, we get $\begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*}$ Then using $[XYZ] = 2R^2\sin X\sin Y\sin Z$ , we can compute $[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.$ Thus $[AEF] = 15\sqrt{7}/14$ , which gives a final answer of $\boxed{036}$ .

~pinetree1

Solution 3(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 4 (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 5 (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 6 (geometry+trig)

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(A--(5,sqrt(28))); draw(A--(0,9sqrt(7)/7)); draw(D--(0,9sqrt(7)/7)); draw(D--(5,sqrt(28))); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$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 therefore these two triangles are not congruent, we know that $\angle BAE$ and $\angle BDE$ are supplementary. Then we know that $\angle ABD$ and $\angle AED$ are also supplementary. 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

Remark: I didn't figure out how to add segments $AF$ , $AE$ , $DF$ and $DE$ . Can someone please help add these segments?

(Added :) ~Math_Genius_164)

Solution 7

First and foremost $\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]$ as $EF$ is the perpendicular bisector of $AD$ . Now note that quadrilateral $ABDF$ is cyclic, because $\angle{ABF}=\angle{FBD}$ and $FA=FD$ . Similarly quadrilateral $AEDC$ is cyclic, $\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}$ Let $A'$ , $B'$ , $C'$ be the $A$ , $B$ , and $C$ excenters of $\triangle{ABC}$ respectively. Then it follows that $\triangle{DEF} \sim \triangle{A'C'B'}$ . By angle bisector theorem we have $BD=2 \implies \dfrac{ID}{IA}=\dfrac{BD}{BA}=\dfrac{1}{2}$ . Now let the feet of the perpendiculars from $I$ and $A'$ to $BC$ be $X$ and $Y$ resptively. Then by tangents we have $BX=s-AC=\dfrac{3}{2} \implies XD=2-\dfrac{3}{2}=\dfrac{1}{2}$ $CY=s-AC \implies YD=3-\dfrac{3}{2}=\dfrac{3}{2} \implies \dfrac{ID}{DA'}=\dfrac{XD}{YD}=\dfrac{1}{3} \implies \big[\triangle{DEF}\big]=\dfrac{1}{16}\big[\triangle{A'C'B'}\big]$ From the previous ratios, $AI:ID:DA'=2:1:3 \implies AD=DA' \implies \big[\triangle{ABC}\big]=\big[\triangle{A'BC}\big]$ Similarly we can find that $\big[\triangle{B'AC}\big]=2\big[\triangle{ABC}\big]$ and $\big[\triangle{C'AB}\big]=\dfrac{4}{7}\big[\triangle{ABC}\big]$ and thus $\big[\triangle{A'B'C'}\big]=\bigg(1+1+2+\dfrac{4}{7}\bigg)\big[\triangle{ABC}\big]=\dfrac{32}{7}\big[\triangle{ABC}\big] \implies \big[\triangle{DEF}\big]=\dfrac{2}{7}\big[\triangle{ABC}\big]=\dfrac{15\sqrt{7}}{14} \implies m+n+p = \boxed{036}$ -tkhalid

Solution 8 -Trigonometry(only)

Trig values we use here:

$\cos A = \frac{9}{16}$

$\cos \frac{A}{2} = \frac{5}{4\sqrt2}$

$\sin \frac{A}{2} = \frac{\sqrt7}{4\sqrt2}$

$\cos \frac{B}{2} = \frac{3}{4}$

$\cos \frac{C}{2} = \frac{\sqrt7}{2\sqrt2}$

First let the incenter be $I$ . Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of $M$ to $BC$ .

We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$ . Then it is easy to find that $AD = 3\sqrt3$ .

Now we trig bash for $DI = MI - MD$ . Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \frac{BK}{\cos \frac{A}{2}} = \frac{\frac{5}{2}}{\frac{5}{4\sqrt2}}=2\sqrt2$ . To get $MD$ we angle chase to get $\angle KDM = \frac{A}{2}+C$ . Then $\cos(\frac{A}{2}+C) = \cos\frac{A}{2}\cos C - \sin\frac{A}{2}\sin C = \frac{1}{2\sqrt2} = \frac{\frac{1}{2}}{MD}$ gives $MD = \sqrt{2}$ . This means $DI = \sqrt2$ .

Now let $AI \cap EF = G$ . It is easy to angle chase $\angle GIE = 90- \frac{B}{2}$ and $\angle GIF = 90- \frac{C}{2}$ . Since $GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}$ , we compute that $EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}$ which implies $[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}$ which gives an answer of $\boxed{36}$ . ~Leonard_my_dude~

Solution 9 (Official MAA 1)

Let $x = \angle BAD = \angle CAD$ , $y = \angle CBE = \angle ABE$ , and $z = \angle BCF = \angle ACF$ . Notice that $x+y+z = 90^\circ$ .

In $\triangle ABD$ , segment $\overline{BE}$ is the bisector of $\angle ABD$ , and $E$ lies on the perpendicular bisector of side $\overline{AD}$ . Therefore $E$ is the midpoint of arc $\stackrel{\textstyle\frown}{AD}$ on the circumcircle of $\triangle ABD$ . It follows that $\angle BED = \angle BAD = x$ and $\angle EDA = \angle EBA = y$ . Likewise, $ACDF$ is cyclic, $\angle CFD = \angle CAD = x$ , and $\angle FDA = \angle FCA = z$ . Because $\overline{EF}$ is the perpendicular bisector of $\overline{AD}$ , triangles $AEF$ and $DEF$ are congruent, implying that $\begin{align*} [\triangle AEF] = [\triangle DEF] &= \frac{DE\cdot DF\cdot\sin(\angle EDF)}{2}\\ &= \frac{DE\cdot DF\cdot\sin(y+z)}{2} = \frac{DE\cdot DF\cdot\cos x}{2}. \end{align*}$ $[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); [/asy]$

Applying the Law of Sines to $\triangle BED$ and $\triangle CFD$ gives $DE = BD\cdot\frac{\sin y}{\sin x}\text{~ and ~} DF = CD\cdot\frac{\sin z}{\sin x}.$ By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$ . Combining the above information yields $[\triangle AEF] = \frac{3\sin y\cdot\sin z\cdot\cos x}{\sin^2 x}.$ Applying the Law of Cosines to $\triangle ABC$ gives $\cos 2x = \frac9{16}$ , $\cos 2y = \frac1{8}$ , and $\cos 2z = \frac34$ . By the Half Angle Formulas, $\sin^2x = \frac7{32},~~ \cos x = \sqrt{\frac{25}{32}},~~ \sin y = \sqrt{\frac7{16}}, \text{~ and ~} \sin z = \sqrt{\frac18}.$ Therefore $[\triangle AEF] = \frac{3\cdot\sqrt{\frac{7}{16}}\cdot\sqrt{\frac{1}{8}}\cdot\sqrt{\frac{25}{32}}} {\frac{7}{32}} = \frac{15\sqrt{7}}{14}.$ The requested sum is $15+7+14 = 36$ .

Solution 10 (Official MAA 2)

Let the point $M$ be the midpoint of $\overline{AD}$ , let $I$ be the incenter of $\triangle ABC$ which is the common point of lines $AD$ , $BE$ , and $CF$ , and let $r$ be the inradius of $\triangle ABC$ . The semiperimeter of $\triangle ABC$ is $s = \frac{AB + BC + CA}2 = \frac{15}2,$ and Heron's Formula gives the area of $\triangle ABC$ as $\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.$ This area is also $rs$ implying that $r = \frac{\sqrt7}2$ . Stewart's Theorem gives $AD =3\sqrt2$ . Because the ratio of the areas of $\triangle IBC$ and $\triangle ABC$ is $\frac{ID}{AD},$ it follows that $ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.$ Thus $IM = MD - ID = \frac{\sqrt2}2$ .

$[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); dot("$I$", I, dir(120)); dot("$M$", M, W); [/asy]$

Note that $\angle EFI = 90^{\circ} - \angle FIA = 90^{\circ} - \angle CID = 90^{\circ} - \frac{\angle A}{2} - \frac{\angle C}{2} = \frac{\angle B}{2} = \angle IBC$ . Thus $\triangle IBC \sim \triangle IFE$ . The height of $\triangle IBC$ to $I$ is $r=\frac{\sqrt7}2$ , and the height of $\triangle IFE$ to $I$ is $IM=\frac{\sqrt2}2$ , so $EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}$ . The needed area of $\triangle AEF$ is $\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}$ , as above.

Solution 11 Bash for life

Firstly, it is easy to find $BD=2,CD=3$ with angle bisector theorem.

Using LOC and some trig formulas we get all those values: $cos\angle{B}=\frac{1}{8},sin\angle{B}=\frac{3\sqrt{7}}{8},cos\angle{\frac{B}{2}}=\frac{3}{4},cos\angle{\frac{{ACB}}{2}}=\frac{\sqrt{14}}{4}$ Now we find the coordinates of points $A,E,F$ and we apply shoelace theorem later. Point A's coordinates is $(4*\frac{1}{8},4*\frac{3\sqrt{7}}{8})=(\frac{1}{2},\frac{3\sqrt{7}}{2})$ , Let $AJ$ is perpendicular to $BC$ , $tan\angle{JAD}=\frac{2-\frac{1}{2}}{\frac{3\sqrt{7}}{2}}=\frac{\sqrt{7}}{7}$ , which means the slope of $FE$ is $\frac{\sqrt{7}}{7}$ . Find the coordinate of $M$ , it is easy, $(\frac{5}{4},\frac{3\sqrt{7}}{4})$ , the function $EF$ is $y=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ . Now find the intersection of $EF,EB$ , $\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}=\frac{\sqrt{7}x}{3}$ , getting that $x=3,E(3,\sqrt{7})$ Now we look at line segment $CF$ , since $cos\angle{\frac{ACB}{2}}=\frac{\sqrt{14}}{4},tan\angle{\frac{ACB}{2}}=\frac{\sqrt{7}}{7}$ . Since the line passes $(5,0)$ , we can set the equation to get the $CF:y=-\frac{\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$ , find the intersection of $CF,EF$ , $-\frac{\sqrt{7}x}{7}+\frac{5\sqrt{7}}{7}=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ , getting that $F:(\frac{1}{2},\frac{9\sqrt{7}}{14})$ and in the end we use shoelace theorem with coordinates of $A,F,E$ getting the area $\frac{15\sqrt{7}}{14}$ leads to the final answer $\boxed{36}$

~bluesoul

Solution 12 (Simple geometry)

Using the Claim (below) we get $I$ is orthocenter of $\triangle DEF,\angle EDG = \beta,$ $\angle FDG = \gamma.$ So area of $\triangle DEF$ is $[\triangle DEF] =\frac {DG \cdot FE}{2} = \frac {AD^2}{8} (\tan \beta + \tan \gamma).$

Semiperimeter of $\triangle ABC \hspace{10mm} s = \frac{15}{2},$ so the bisector $AD= \frac{2}{AB + AC}\sqrt{s(s-BC) \cdot AB\cdot AC} = 3\sqrt{2}.$

We get the inradius by applying Heron's formula $r = \sqrt{\frac {(s-AB)(s-BC)(s-AC)}{s}} = \sqrt {\frac {3\cdot 5 \cdot 7}{15 \cdot 4}}=\frac {\sqrt {7}}{2}.$

We use formulas for inradius and get $\tan \beta = \frac{r}{s – AC} = \frac {\sqrt {7}}{3} , \hspace{10mm} \tan \gamma = \frac{r}{s – AB} = \frac {\sqrt {7}}{7}.$ The area $\hspace{30mm} [\triangle DEF] = \frac{15 \sqrt 7}{14}.$

Claim

Let $I$ be incenter of $\triangle ABC.$ Then bisector $\overline{BI},$ perpendicular bisector of $\overline{AD},$ and perpendicular dropped to bisector $\overline{CI}$ from point $D$ are concurrent.

Proof

Denote $\angle BAC = 2 \alpha, \angle ABC = 2\beta,\angle ACB = 2 \gamma.$ Then $\alpha + \beta + \gamma = 90^\circ.$ Denote $P$ the intersection point of $BC$ and the tangent line to the circumcircle at point $A.$

WLOC, $\hspace{20mm}\gamma > \beta$ (case $\gamma = \beta$ is trivial).

$\angle PAC = \angle ABC = 2 \beta$ (this angles are measured by half the arc $\overset{\Large\frown} {AC}$ of the circumcircle).

$\angle APC = \angle ACD - \angle PAC = 2(\gamma - \beta),$ $\angle ADP = 180^o – \angle DAC - \angle ACD = 180^o – \alpha – 2 \gamma = \alpha + 2 \beta = \angle DAP \implies AP = DP.$ Therefore bisector of angle P coincite with the perpendicular bisector of $\overline{AD}$ .

By applying the Law of Sines to $\triangle ABP$ we get $\hspace{20mm}\frac {BP}{AP} = \frac {\sin 2 \gamma}{\sin 2 \beta.}$

Let $E$ be crosspoint of $PG$ and bisector $BI.$ By applying the Law of Sines to $\triangle BEP$ we get $\frac {EP}{BP} = \frac {\sin \beta}{\sin(180^o – \beta – (\gamma – \beta))}= \frac {\sin \beta}{\sin \gamma}.$

Let $E'$ be crosspoint of $PG$ and the perpendicular dropped to bisector $\overline{CI}$ from point $D.$ $\frac {E'P}{DP} = \frac {\sin (90^\circ – \gamma)}{\sin(180^o - (90^o – \gamma) – (\gamma - \beta))}= \frac {\cos\gamma}{\cos \beta}.$ $\hspace{50mm}\frac {E'P} {EP} = \frac {E'P}{DP} \cdot \frac {AP}{BP} \cdot \frac {BP}{EP} = \frac {\cos\gamma}{\cos \beta} \cdot \frac {\sin 2\beta}{\sin 2\gamma}\cdot \frac {\sin\gamma}{\sin \beta} = 1 \implies E$ coincide with $E'.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/NpB7zpy-yKM

~MathProblemSolvingSkills.com

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
All AIME Problems and Solutions

Page

Toolbox

Search