1950 AHSME Problems/Problem 47
Problem
A rectangle inscribed in a triangle has its base coinciding with the base 
of the triangle. If the altitude of the triangle is 
, and the altitude 
of the rectangle is half the base of the rectangle, then:
Solution 1:Similarity
Draw the triangle.The small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle.Hence, the proportions of the heights is equal to the proportions of the sides.
In particular, we get
![\[\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}\]](http://latex.artofproblemsolving.com/b/8/e/b8e162c19ea6bac23c6f451167fd3f793d733d5c.png)
The answer is 
.
Solution 2: Area
The smaller triangle at the top has an area of ![\[\dfrac{2x(h-x)}{2}=x(h-x)\]](http://latex.artofproblemsolving.com/8/4/1/8419f30afd399d191c951af0165adb58b840dddb.png)
Because the two other pieces are complementary triangles, we can add them together to create one triangle with area
![\[\dfrac{x(b-2x)}{2}\]](http://latex.artofproblemsolving.com/1/c/9/1c945d99f919f2095d7c854b55261e95ba7e5fb6.png)
Lastly the area of the rectangle is 
. These areas together sum to the area of the big rectangle, which is 
. Hence,
![\[\dfrac{bh}{2}=2x^2 + dfrac{x(b-2x)}{2}+x(h-x)\]](http://latex.artofproblemsolving.com/b/e/d/bed6ee63fe26352ecbafbfefae346c2eb8c4d20d.png)
Solving we get that ![\[x = \dfrac{bh}{2h + b}\]](http://latex.artofproblemsolving.com/f/8/9/f89677f637c88dad05fc1ac35fd7d49b34134bc4.png)
The answer is .
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=169s
~MathProblemSolvingSkills.com
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
