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

(→‎Solution 2(coordinate bash based))
(→‎Solution 1)
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Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{12}{\sqrt{7}}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11
 
Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{12}{\sqrt{7}}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11
 +
 +
==Solution 2==
 +
Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=3\sqrt{7}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (3\sqrt{7}{2})^2=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>.
 +
Since we want the area of triangle <math>AEF</math>, 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 <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>.
 +
The slope of AB, or in trig words, the tangent of <math>\angle ABC</math> is <math>3\sqrt{7}</math>.
 +
Finding <math>\sin{\angle ABC}=\frac{\frac{3\sqrt{7}}{2}}{4}=\frac{3\sqrt{7}}{8}</math> and <math>\cos{\angle ABC}=\frac{\frac{1}{2}}{4}=\frac{1}{8}</math>. Plugging this in to half angle tangent, it gives <math>\frac{\frac{3\sqrt{7}}{8}}{1+\frac{1}{8}}=\frac{\sqrt{7}}{3}</math> as the slope of the angle bisector, since it passes through <math>B</math>, the equation is <math>y=\frac{\sqrt{7}}{3}x</math>.
 +
Similarly, the equation for the angle bisector of <math>C</math> will be $y=-\frac{1}{\sqrt{7}}(x-5).
  
 
==See Also==
 
==See Also==

Revision as of 16:58, 12 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 $m+n+p.$


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

See Also

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