2010 aime #12

**Mrdavid445** · December 30th 2012, 10:52 AM

Two non-congruent inter-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.

When I did, I got a very ugly radical, but AIME answers are always whole numbers. Solution please?

**ILikeSerena** · December 30th 2012, 02:43 PM

Originally Posted by Mrdavid445

Two non-congruent inter-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.

When I did, I got a very ugly radical, but AIME answers are always whole numbers. Solution please?

Hi Mrdavid445!

I don't know what inter-sided means.
Can you clarify?

Either way, whatever the minimum value of the common perimeter is, when you make those triangles twice as small, doesn't the perimeter become twice as small as well?
What am I misunderstanding?

**Wilmer** · December 30th 2012, 05:17 PM

Triangles are: 233-233-210 and 218-218-240

Perimeter = 676, area = 21840
240:210 = 8:7

Best way is stick together 2 pythagorean triangles to get an isosceles triangle:
like 233-233-210 from 2 "105-208-233", thus 208 = height.

NOTE that I'm assuming that (somehow!) your problem means:
2 isoceles triangles with integer sides and different bases have the same integer area and integer perimeter.
The ratio of the bases is 8:7. Find the smallest case.

Like Serena, I've never heard of "inter-sided" triangles!

EDIT: unable to add a new post...(wonder why)...so I'll go this way:
on Soroban's solution, a bit easier if 7k is used as height instead of h, and 8k instead of k

**Soroban** · December 30th 2012, 06:54 PM

Hello, Mrdavid445!

Two non-congruent 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.

The two triangles looks like this:

Code:

                                   *
              *                   *|*
         a  * | *  a           b * | * b
          *   |h  *             *  |k *
        *     |     *          *   |   *
      *-------+-------*       *----+----*
      : - -  8c - - - :       : - 7c  - :

Triangle $A$ has equal sides $a$ , base $8c$ , and altitude $h \,=\,\sqrt{a^2 - 4c^2}$

Triangle $B$ has equal sides $b$ , base $7c$ and altitude $k \,=\,\sqrt{b^2 + \tfrac{49}{4}c^2}$

Their perimeters are equal: . $2a + 8c \:=\:2b + 7c \quad\Rightarrow\quad b \,=\,a+ \tfrac{c}{2}$ .[1]

Their areas are equal: . ${\color{red}\rlap{/}}\tfrac{1}{2}(8{\color{red}\rlap{/}}c)\sqrt{a^2-16c^2} \:=\:{\color{red}\rlap{/}}\tfrac{1}{2}(7{\color{red}\rlap{/}}c)\sqrt{b^2-\tfrac{49}{4}c^2}$

Square both sides: . $64(a^2-16c^2) \;=\;49\left(b^2-\tfrac{49}{4}c^2\right)$

n . . . . . . . . . . . . . $64a^2 - 1024c^2 \;=\;49b^2 - \tfrac{2401}{4}c^2$

. . . . . . . . . . . . . . . . $64a^2 - 49b^2 \;=\;\tfrac{1695}{4}c^2$

. . . . . . . . . . . . . . . $256a^2 - 196b^2 \;=\;1695c^2$

Substitute [1]: . $256a^2 - 196\left(a+\tfrac{c}{2}\right)^2 \;=\;1695c^2$

n . . . . . . $256a^2 - 196\left(a^2 + ac + \tfrac{c^2}{4}\right) \;=\;1695c^2$

n . . . . $256a^2 - 196a^2 - 196ac - 49c^2 \;=\;1695c^2$

n . . . . . . . . . $60a^2 - 196ac - 1744c^2 \;=\;0$

. . . . . . . . . . . . $15a^2 - 49ac - 436c^2 \;=\;0$

n . . . . . . . . . . $(15a - 109c)(a + 4c) \;=\;0$

Hence: . $15a - 109c \:=\:0 \quad\Rightarrow\quad a \:=\:\tfrac{109}{15}c$

Substitute into [1]: . $b \:=\:\tfrac{109}{15}c+\tfrac{1}{2}c \quad\Rightarrow\quad b\:=\:\tfrac{233}{30}c$

For the least integral dimensions, let $c = 30$

Hence: . $a = 218,\;b = 233$

Perimeter of triangle $A$ is: . $2a + 8c \:=\:2(218) + 8(30) \:=\:676$
Perimeter of triangle $B$ is: . $2b + 7c \:=\:2(233) + 7(30) \:=\:676$

**skeeter** · January 1st 2013, 10:05 AM

Reply working?

**Wilmer** · January 1st 2013, 11:35 AM

Seems to be...but definitely wasn't...

**ILikeSerena** · January 1st 2013, 02:03 PM

Originally Posted by Wilmer

Seems to be...but definitely wasn't...

Hmm, you are not the OP, are you?

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