2010 AIME II Problems/Problem 12

Problem

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter.

Solution 1

Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have

$\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}$ $8\sqrt{s-16x}=7\sqrt{s-14x}$ $64s-1024x=49s-686x$ $15s=338x$

Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$ . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$ , which gives us a final answer of $\boxed{676}$ .

Solution 2

Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have $a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).$ Multiplying this, we get $64a^2-4096n^2=49a^2+98an-2352n^2,$ which simplifies to $15a^2-98an-1744n^2=0.$ Solving this for $a$ , we get $a=n\cdot\frac{218}{15}$ , so $n=15$ and $a=218$ and the perimeter is $15\cdot16+218+218=\boxed{676}$ .

~john0512

Note

We use $16x$ and $14x$ instead of $8x$ and $7x$ to ensure that the triangle has integral side lengths. Plugging $8x$ and $7x$ directly into Heron's gives $s=338$ , but for this to be true, the second triangle would have side lengths of $\frac{223}{2}$ , which is impossible.

~jd9

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2010 AIME II Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

Note

See also