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

But try the problem first…

Answer: is 108.

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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