Difference between revisions of "2010 AIME II Problems/Problem 12"
Mathgeek2006 (talk | contribs) m (→Solution 1) |
m (→Solution 1) |
||
Line 22: | Line 22: | ||
<math>\begin{array}{cccl} | <math>\begin{array}{cccl} | ||
7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\ | 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\ | ||
− | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\ | + | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{(b+8c)(b-8c)})&{}\ |
7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\ | 7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\ | ||
49a-343c&=&64b-512c&{}\ | 49a-343c&=&64b-512c&{}\ |
Revision as of 16:00, 21 January 2016
Contents
[hide]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 . Find the minimum possible value of their common perimeter.
Solution 1
Let the first triangle have side lengths ,
,
, and the second triangle have side lengths
,
,
, where
.
Equal perimeter:
Equal Area:
Since and
are integer, the minimum occurs when
,
, and
. Hence, the perimeter is
.
Solution 2
Let be the semiperimeter of the two triangles. Also, let the base of the longer triangle be
and the base of the shorter triangle be
for some arbitrary factor
. Then, the dimensions of the two triangles must be
and
. By Heron's Formula, we have
Since and
are coprime, to minimize, we must have
and
. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by
, which gives us a final answer of
.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.