2021 Fall AMC 10B Problems/Problem 13
A square with side length is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
Let's split the triangle down the middle and label it:
We see that by AA similarity. because cuts the side length of the square in half; similarly, . Let : then by side ratios,
Now the height of the triangle is . By side ratios, .
The area of the triangle is
By similarity, the height is and the base is . Thus the area is , or .
~Hefei417, or 陆畅 Sunny from China
Solution 3 (With two different endings)
This solution is based on this figure: Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png
Denote by the midpoint of .
Because , , , we have .
We observe . Hence, . Hence, . By symmetry, .
Because is the midpoint of , .
We observe . Hence, . Hence, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Alternatively, we can find the height in a slightly different way.
Following from our finding that the base of the large triangle , we can label the length of the altitude of as . Notice that . Hence, . Substituting and simplifying, . Therefore, the area of the triangle is .
Video Solution by Interstigation
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