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

Revision as of 20:32, 18 August 2017

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 (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 2 does.

Solution 2 (Trig)

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 $2(x+y) = \boxed{108}$ .

@@ Line 27: / Line 27: @@
 Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
-==Solution 3 (More Trig)== -ccx09 (Someone please help clean this up. Thanks)
-Let x be the measure of <ABI. Thus, <ABI=<CBI=<ACI=<BCI=x, and x<45 degrees. Note that the angle bisector of <BAC will pass through I and since AB=AC, will also be the perpendicular bisector of BC. Call the point of intersection of segments AI and BC point M.
-Let BM=CM=a, and AB=AC=b. Thus, we want to minimize p=2(a+b), or minimize a+b.
-We now consider triangle BAM. We note that cos(x)=BM/BI=a/8, and cos(2x)=a/b.
-We can now simplify as follows. cos(2x)=2(cos x)(cos x)-1, and since cos(x)=a/8, we have
-a/b = 2(a/8)^2-1 = a^2/32-1 = (a^2-32)/32. Therefore b=32a/(a^2-32).
-And thus a + b = a + 32a / (a^2-32). Substituting 8cos(x) for a, we obtain
-a + b = 8cos(x) + 32(8cos x) / ((8cos x)^2)-32).
-Since a and b are integers, we must have cos(x) = 0, 1/8, 2/8, 3/8, 4/8, 5/8. 6/8, 7/8, or 1.
-However, since x<45 degrees, we must have cos(x)>=6/8. To minimize a + b, we thus let cos(x)=6/8. Plugging cos(x)=6/8 into the previous expression of a + b yields a + b = 54.
-This is the minimum possible value of a + b.
-Thus, the minimum perimeter of triangle ABC is 54*2 = 108.
 ==See Also==

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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Difference between revisions of "2015 AIME I Problems/Problem 11"

Revision as of 20:32, 18 August 2017

Contents

Problem

Solution 1 (No Trig)

Solution 2 (Trig)

See Also