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Smallest Perimeter of Triangle | AIME I, 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

Smallest Perimeter of Triangle – AIME 2015


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 \(\triangle ABC\)..

  • is 107
  • is 108
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequalities

Trigonometry

Geometry

Check the Answer


Answer: is 108.

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

Try with Hints


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 $\angle IBD$ =$ \frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} – 1 = \frac{y^2-32}{32}$.

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 $x=\frac{32(6)}{(6)^2-32}=48$.

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

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