2015 AMC 12A Problems/Problem 20
Problem
Isosceles triangles and are not congruent but have the same area and the same perimeter. The sides of have lengths , , and , while those of have lengths , , and . Which of the following numbers is closest to ?
Solution
Solution 1
The area of is and the perimeter is 18.
The area of is and the perimeter is .
Thus , so .
Thus , so .
We square and divide 36 from both sides to obtain , so . This factors as . Because clearly but , we have The answer is .
Solution 2
Triangle , being isosceles, has an area of and a perimeter of . Triangle similarly has an area of and .
Now we apply our computational fortitude.
Plug in to obtain Plug in to obtain We know that is a valid solution by . Factoring out , we obtain Utilizing the quadratic formula gives We clearly must pick the positive solution. Note that , and so , which clearly gives an answer of , as desired.
Solution 3
Triangle T has perimeter so .
Using Heron's, we get .
We know that from above so we plug that in, and we also know that then .
We plug in 3 for in the LHS, and we get 54 which is too low. We plug in 4 for in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4.
If , then we would round up to 4, but if , then we would round down to 3. So let us plug in 3.5 for b.
We get 67.375 which is too high, so we know that .
The answer is .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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All AMC 12 Problems and Solutions |