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Difference between revisions of "2020 AIME I Problems/Problem 13"

m (→‎Solution 2)
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The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
 
The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. 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 <math>B=(0,0)</math> and <math>BC</math> be the line <math>y=0</math>.
 +
We compute that <math>\cos{\angle{ABC}}=\frac{1}{8}</math>, so <math>\tan{\angle{ABC}}=3\sqrt{7}</math>.
 +
Thus, <math>A</math> lies on the line <math>y=3x\sqrt{7}</math>. The length of <math>AB</math> at a point <math>x</math> is <math>8x</math>, so <math>x=\frac{1}{2}</math>.
 +
 +
We now have the coordinates <math>A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)</math>, <math>B=(0,0)</math> and <math>C=(5,0)</math>.
 +
We also have <math>D=(2,0)</math> by the angle-bisector theorem and <math>M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)</math> by taking the midpoint.
 +
We have that because <math>\cos{\angle{ABC}}=\frac{1}{8}</math>, <math>\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}</math> by half angle formula.
 +
 +
We also compute <math>\cos{\angle{ACB}}=\frac{3}{4}</math>, so <math>\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}</math>.
 +
 +
Now, <math>AD</math> has slope <math>-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}</math>, so it's perpendicular bisector has slope <math>\frac{\sqrt{7}}{7}</math> and goes through <math>\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)</math>.
 +
 +
We find that this line has equation <math>y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}</math>.
 +
 +
As <math>\cos{\angle{CBI}}=\frac{3}{4}</math>, we have that line <math>BI</math> has form <math>y=\frac{\sqrt{7}}{3}x</math>.
 +
Solving for the intersection point of these two lines, we get <math>x=3</math> and thus <math>E=\left(3, \sqrt{7}\right)</math>
 +
 +
We also have that because <math>\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}</math>, <math>CI</math> has form <math>y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}</math>.
 +
 +
Intersecting the line <math>CI</math> and the perpendicular bisector of <math>AD</math> yields <math>-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}</math>.
 +
 +
Solving this, we get <math>x=\frac{1}{2}</math> and so <math>F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)</math>.
 +
 +
We now compute <math>EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}</math>.
 +
We also have <math>MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}</math>.
 +
 +
As <math>{MA}\perp{EF}</math>, we have <math>[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}</math>.
 +
 +
 +
The desired answer is <math>15+7+14=\boxed{036}</math> ~Imayormaynotknowcalculus
 
==See Also==
 
==See Also==
  
 
{{AIME box|year=2020|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2020|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:51, 14 March 2020

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

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

See Also

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

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