AIME 2010 II · 第 12 题

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

Solution 3

Let the sides of the first triangle be ${a, a, 8k}$ and the sides of the second triangle be ${b, b, 7k}$

Aim: We want to create some sort of relation between the perimeter and area since those elements are common in both triangles.

First to find the perimeter of both triangles we do

$P = 2a + 8k$ where P is the perimeter of the first triangle $P = 2b + 8k$ where P is the perimeter of the second triangle

Both equations use the same variable P since the perimeters of the triangles are equal as stated in the problem.

Next to find the area of both triangles we multiply the height and base and divide by 2.

To get the height of the first triangle we do pythag: $\sqrt{a^2 - (4k)^2}$ Do the same for the second triangle and you get $\sqrt{b^2 - (7k/2)^2}$

Multiple both by there bases and divide by 2 to get Area of 1st triangle = $(8k)^2*(\sqrt{4a^2-(8k)^2}$ Area of 2nd triangle = $(7k)^2*(\sqrt{4b^2-(7k)^2}$

Since both areas are equal we can create an equation:

$(8k)^2*(\sqrt{4a^2-(8k)^2}$ = $(7k)^2*(\sqrt{4b^2-(7k)^2}$

Square both sides and divide by k^2:

$64*(4a^2-64k^2) = 49(4c^2-49k^2)$

Expand and combine like terms to get:

$$256a^2 - 196c^2 = 1695k^2$$ This factors into $(16a + 14c)(16a - 14c) = 1695k^2$

recall that P = 2a + 8k and P = 2c + 7k, substituting those values gives

$$(P - 15k)(15P - 113k) = 1695k^2$$ Expand and combine like terms to get:

$$15P^2 - 338Pk = 0$$ We can further simplify this to:

$P = \frac{338k}{15}$

Proceed from this point like the other solutions!

Notes

The triangles in question have sides $\boxed{240,218,218}$ and $\boxed{210,233,233}$. ~eevee9406

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