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

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Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.
 
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.
  
−
==Solution 1 (No Trig)==
+
==Solution 1==
−
Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By pythagorean theorem, <math>AE=\sqrt{x^2+y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By angle bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2+y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using pythagorean on <math>BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as solution 2 does.
 
  
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Note: In solution 2, the variables for <math>x</math> and <math>y</math> are switched.
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Let <math>D</math> be the midpoint of <math>\overline{BC}</math>. Then by SAS Congruence, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle ADB = \angle ADC = 90^o</math>.
 +
 
 +
Now let <math>BD=y</math>, <math>AB=x</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>.
 +
 
 +
Then <math>\mathrm{cos}{(\theta)} = \dfrac{y}{8}</math>
 +
 
 +
and <math>\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}</math>.
 +
 
 +
Cross-multiplying yields <math>32y = x(y^2-32)</math>.
  
−
==Solution 2 (Trig)==
+
Since <math>x,y>0</math>, <math>y^2-32</math> must be positive, so <math>y > 5.5</math>.
  
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Let <math>D</math> be the midpoint of <math>\overline{BC}</math>. Then by SAS Congruence, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle ADB = \angle ADC = 90^o</math>.
+
Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=y < 8</math>.
  
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Now let <math>BD=x</math>, <math>AB=y</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>.
+
Therefore, given that <math>BC=2y</math> is an integer, the only possible values for <math>y</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>.
  
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Then <math>\mathrm{cos}{(\theta)} = \dfrac{x}{8}</math>
+
However, only one of these values, <math>y=6</math>, yields an integral value for <math>AB=x</math>, so we conclude that <math>y=6</math> and <math>x=\dfrac{32(6)}{(6)^2-32}=48</math>.
  
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and <math>\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32}</math>.
+
Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
  
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Cross-multiplying yields <math>32x = y(x^2-32)</math>.
+
==Solution 2 (No Trig)==
 +
Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By Pythagorean Theorem, <math>AE=\sqrt{x^2-y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By Angle Bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2-y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using Pythagorean Theorem on <math>\triangle BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as Solution 1 does.
  
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Since <math>x,y>0</math>, <math>x^2-32</math> must be positive, so <math>x > 5.5</math>.
+
==Solution 3==
 +
Let <math>AB=x</math>, call the midpoint of <math>BC</math> point <math>E</math>, call the point where the incircle meets <math>AB</math> point <math>D</math>,
  
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Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=x < 8</math>.
+
and let <math>BE=y</math>. We are looking for the minimum value of <math>2(x+y)</math>. <math>AE</math> is an altitude because the triangle
 +
 +
is isosceles. By Pythagoras on <math>BEI</math>, the inradius is <math>\sqrt{64-y^2}</math> and by Pythagoras on <math>ABE</math>, <math>AE</math> is
  
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Therefore, given that <math>BC=2x</math> is an integer, the only possible values for <math>x</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>.
+
<math>\sqrt{x^2-y^2}</math>. By equal tangents, <math>BE=BD=y</math>, so <math>AD=x-y</math>. Since <math>ID</math> is an inradius, <math>ID=IE</math> and
 +
using pythagoras on <math>ADI</math> yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we
  
−
However, only one of these values, <math>x=6</math>, yields an integral value for <math>AB=y</math>, so we conclude that <math>x=6</math> and <math>y=\dfrac{32(6)}{(6)^2-32}=48</math>.
+
can write <math>\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}</math>. Simplifying, <math>\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}</math>.  
  
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Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
+
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 <math>2 * 48 + 12 = \boxed{108}</math>.
 +
- whatRthose
  
 
==See Also==
 
==See Also==

Revision as of 16:08, 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 $2 * 48 + 12 = \boxed{108}$. - whatRthose

See Also

2015 AIME I (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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