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 Solution 13 (Trig)
16 Video Solution
17 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

Solution 13 (Trig)

Denote $I$ as the incenter of $\triangle ABC$ and $M$ as the intersecting point of $AD$ and $EF$ .

We solve this question based on this equation: $[\triangle AEF]=\dfrac{1}{2}\times AM\times (\dfrac{IM}{\tan{\angle EFI}}+\dfrac{IM}{\tan{\angle FEI}})$ .

Firstly, let's compute the trig value of $\angle EFI$ and $\angle FEI$ . Using the Cosine Law and half-angle formula, we obtain: $\cos{\angle B}=\dfrac{1}{8}, \sin{\dfrac{\angle B}{2}}=\dfrac{\sqrt{7}}{4}, \cos{\dfrac{\angle B}{2}}=\dfrac{3}{4}$ $\cos{\angle C}=\dfrac{3}{4}, \sin{\dfrac{\angle C}{2}}=\dfrac{1}{\sqrt{8}}, \cos{\dfrac{\angle C}{2}}=\dfrac{\sqrt{7}}{\sqrt{8}}$

Noticing that by angle chasing, $\angle EFI= 90^{\circ}-\angle FIA=90^{\circ}-\angle CID=90^{\circ}-(180^{\circ}-\dfrac{\angle A}{2}-\angle B-\dfrac{\angle C}{2})= \dfrac{\angle B}{2}$ . Similarly, $\angle FEI=\dfrac{\angle C}{2}$ . Thus, $\tan{\angle EFI}=\dfrac{\sqrt{7}}{3}$ and $\tan{\angle FEI}=\dfrac{1}{\sqrt{7}}$

Then we find $AM$ : by Stewart's theorem, $AM=\dfrac{AD}{2}=\dfrac{3\sqrt{2}}{2}$ .

For $MI$ , first noticing that $MI=MD-ID=\dfrac{3\sqrt{2}}{2}-ID$ . Then, by Angel Bisector's theorem, we have $BD=2, CD=3$ , and applying Sine Law: $\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\sin{\angle CID}}$ $\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\cos{\dfrac{\angle B}{2}}}$ Substituting the known values, we get $ID=\sqrt{2}$ . Thus, $MI=\dfrac{\sqrt{2}}{2}$

Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain $[\triangle AEF]=\dfrac{15\sqrt{7}}{14}$ , which gives a final answer of $\boxed{036}$ .

~Ericcc

Video Solution

https://youtu.be/NpB7zpy-yKM

~MathProblemSolvingSkills.com

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