2021 Fall AMC 10B Problems/Problem 13
Contents
Problem
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?
Solution 1
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
~KingRavi
Solution 2
By similarity, the height is and the base is . Thus the area is .
~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, .
Therefore, .
Because is the midpoint of , .
We observe . Hence, . Hence, .
Therefore, .
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 .
~mahaler
Solution 4 (Coordinates)
For convenience, we will use the image provided in the third solution.
We can set as the origin.
We know that and .
We subtract from and divide by to get .
Since is a square, we know that .
Using rise over run, we find that the slope of is .
The coordinates of are . We plug this in to get the equation of the line that runs along:
We know that the of is . Using this, we find that the is . So the coordinates of are .
This gives us the height of : .
Now we need to find the coordinates of .
We know that the is . Plugging this in, we find , or .
The coordinates of are .
Since is symmetrical along , we can multiply by to get
Simplifying, we get for the area.
~Achelois
Video Solution by Interstigation
https://www.youtube.com/watch?v=mq4e-s9ENas
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/R7TwXgAGYuw?t=639
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.