Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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~Hefei417, or 陆畅 Sunny from China | ~Hefei417, or 陆畅 Sunny from China | ||
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+ | == Solution 3 == | ||
+ | Denote by <math>O</math> the midpoint of <math>AB</math>. | ||
+ | |||
+ | Because <math>FG = 3</math>, <math>JK = 2</math>, <math>FJ = KG</math>, we have <math>FJ = \frac{1}{2}</math>. | ||
+ | |||
+ | We observe <math>\triangle ADF \sim \triangle FJH</math>. | ||
+ | Hence, <math>\frac{AD}{FJ} = \frac{FD}{HJ}</math>. | ||
+ | Hence, <math>AD = \frac{3}{4}</math>. | ||
+ | By symmetry, <math>BE = AD = \frac{3}{4}</math>. | ||
+ | |||
+ | Therefore, <math>AB = AD + DE + BE = \frac{9}{2}</math>. | ||
+ | |||
+ | Because <math>O</math> is the midpoint of <math>AB</math>, <math>AO = \frac{9}{4}</math>. | ||
+ | |||
+ | We observe <math>\triangle AOC \sim \triangle ADF</math>. | ||
+ | Hence, <math>\frac{OC}{DF} = \frac{AO}{AD}</math>. | ||
+ | Hence, <math>OC = 9</math>. | ||
+ | |||
+ | Therefore, <math>{\rm Area} \ \triangle ABC = \frac{1}{2} AB \cdot OC = \frac{81}{4} = 20 \frac{1}{4}</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) }20 \frac{1}{4}}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:46, 25 November 2021
Contents
[hide]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 , or .
~Hefei417, or 陆畅 Sunny from China
Solution 3
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)
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.