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Difference between revisions of "2015 AIME I Problems/Problem 11"

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Squaring, subtracting 1 from both sides, and multiplying everything out, we get <math>yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2</math>, which turns into <math>32y=x(y^2-32)</math>. Finish as in Solution 1.
 
Squaring, subtracting 1 from both sides, and multiplying everything out, we get <math>yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2</math>, which turns into <math>32y=x(y^2-32)</math>. Finish as in Solution 1.
 +
==Solution 4==
 +
Angle bisectors motivate trig bash.
 +
Define angle <math>IBC = x</math>. Foot of perpendicular from <math>I</math> to <math>BC</math> is point <math>P</math>.
 +
<math>\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N</math>, where <math>N</math> is an integer. Thus, <math>\cos(x) = \frac{N}{16}</math>. Via double angle, we calculate <math>\overline{AB}</math> to be <math>\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}</math>. This is to be an integer.  We can bound <math>N</math> now, as <math>N > 11</math> to avoid negative values and <math>N < 16</math> due to triangle inequality. Testing, <math>N = 12</math> works, giving <math>\overline{AB} = 48, \overline{BC} = 12</math>.
 +
Our answer is $2 * 48 + 12 = \boxed{108})
  
 
==See Also==
 
==See Also==

Revision as of 15:07, 28 September 2019

Problem

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.

Solution 1

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.

Now let $BD=y$, $AB=x$, and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$.

Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$

and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}$.

Cross-multiplying yields $32y = x(y^2-32)$.

Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$.

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\dfrac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = \boxed{108}$.

Solution 2 (No Trig)

Let $AB=x$ and the foot of the altitude from $A$ to $BC$ be point $E$ and $BE=y$. Since ABC is isosceles, $I$ is on $AE$. By Pythagorean Theorem, $AE=\sqrt{x^2-y^2}$. Let $IE=a$ and $IA=b$. By Angle Bisector theorem, $\frac{y}{a}=\frac{x}{b}$. Also, $a+b=\sqrt{x^2-y^2}$. Solving for $a$, we get $a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}$. Then, using Pythagorean Theorem on $\triangle BEI$ we have $y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64$. Simplifying, we have $y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64$. Factoring out the $y^2$, we have $y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64$. Adding 1 to the fraction and simplifying, we have $\frac{y^2x(x+y)}{(x+y)^2}=32$. Crossing out the $x+y$, and solving for $x$ yields $32y = x(y^2-32)$. Then, we continue as Solution 1 does.

Solution 3

Let $AB=x$, call the midpoint of $BC$ point $E$, call the point where the incircle meets $AB$ point $D$,

and let $BE=y$. We are looking for the minimum value of $2(x+y)$. $AE$ is an altitude because the triangle

is isosceles. By Pythagoras on $BEI$, the inradius is $\sqrt{64-y^2}$ and by Pythagoras on $ABE$, $AE$ is

$\sqrt{x^2-y^2}$. By equal tangents, $BE=BD=y$, so $AD=x-y$. Since $ID$ is an inradius, $ID=IE$ and using pythagoras on $ADI$ yields $AI=$$\sqrt{x^2-2xy+64}$. $ADI$ is similar to $AEB$ by $AA$, so we

can write $\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}$. Simplifying, $\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}$.

Squaring, subtracting 1 from both sides, and multiplying everything out, we get $yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2$, which turns into $32y=x(y^2-32)$. Finish as in Solution 1.

Solution 4

Angle bisectors motivate trig bash. Define angle $IBC = x$. Foot of perpendicular from $I$ to $BC$ is point $P$. $\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N$, where $N$ is an integer. Thus, $\cos(x) = \frac{N}{16}$. Via double angle, we calculate $\overline{AB}$ to be $\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}$. This is to be an integer. We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\overline{AB} = 48, \overline{BC} = 12$. Our answer is $2 * 48 + 12 = \boxed{108})

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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