Difference between revisions of "2014 AMC 12A Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
| − | Since the total area of each congruent isosceles triangle is the same, the area of each is <math>{\dfrac{1}3</math> the total area of the equilateral triangle of side length 1, or <math>{\dfrac{1}3</math> x <math>{\dfrac{\sqrt3}4</math>. Likewise, the area of each can be defined as <math>{\dfrac{bh}2</math> with base <math>b</math> equaling 1, meaning that <math>{\dfrac{h}2</math> = <math>{\dfrac{1}3</math> x <math>{\dfrac{\sqrt3}4</math>, or <math>h</math> = <math>{\dfrac{\sqrt3}6</math>. A side length of the isosceles triangle is the hypotenuse with legs <math>{\dfrac{b}2</math> and <math>h</math>. Using the Pythagorean Theorem, the side length is <math>\sqrt{{(\dfrac{1}2)}^2 + {({\dfrac{\sqrt3}6)}^2}</math>, or <math>\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}</math>. | + | Since the total area of each [[congruent]] isosceles triangle is the same, the area of each is <math>{\dfrac{1}3</math> the total [[area of an equilateral triangle|area of the equilateral triangle]] of side length 1, or <math>{\dfrac{1}3</math> x <math>{\dfrac{\sqrt3}4</math>. Likewise, the area of each can be defined as <math>{\dfrac{bh}2</math> with base <math>b</math> equaling 1, meaning that <math>{\dfrac{h}2</math> = <math>{\dfrac{1}3</math> x <math>{\dfrac{\sqrt3}4</math>, or <math>h</math> = <math>{\dfrac{\sqrt3}6</math>. A side length of the isosceles triangle is the hypotenuse with legs <math>{\dfrac{b}2</math> and <math>h</math>. Using the Pythagorean Theorem, the side length is <math>\sqrt{{(\dfrac{1}2)}^2 + {({\dfrac{\sqrt3}6)}^2}</math>, or <math>\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}</math>. |
(Solution by johnstucky) | (Solution by johnstucky) | ||
Revision as of 17:39, 9 February 2014
Contents
Problem
Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length 
. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?
Solution 1
Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or 
.
(Solution by djmathman)
Solution 2
Since the total area of each congruent isosceles triangle is the same, the area of each is ${\dfrac{1}3$ (Error compiling LaTeX. Unknown error_msg) the total area of the equilateral triangle of side length 1, or ${\dfrac{1}3$ (Error compiling LaTeX. Unknown error_msg) x ${\dfrac{\sqrt3}4$ (Error compiling LaTeX. Unknown error_msg). Likewise, the area of each can be defined as ${\dfrac{bh}2$ (Error compiling LaTeX. Unknown error_msg) with base 
equaling 1, meaning that ${\dfrac{h}2$ (Error compiling LaTeX. Unknown error_msg) = ${\dfrac{1}3$ (Error compiling LaTeX. Unknown error_msg) x ${\dfrac{\sqrt3}4$ (Error compiling LaTeX. Unknown error_msg), or 
= ${\dfrac{\sqrt3}6$ (Error compiling LaTeX. Unknown error_msg). A side length of the isosceles triangle is the hypotenuse with legs ${\dfrac{b}2$ (Error compiling LaTeX. Unknown error_msg) and . Using the Pythagorean Theorem, the side length is $\sqrt{{(\dfrac{1}2)}^2 + {({\dfrac{\sqrt3}6)}^2}$ (Error compiling LaTeX. Unknown error_msg), or .
(Solution by johnstucky)
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
