Difference between revisions of "2013 AMC 12A Problems/Problem 11"

Revision as of 22:55, 6 February 2013

Let $AD = x$ , and $AG = y$ . We want to find $DE + FG$ , which is nothing but $x+y$ .

Based on the fact that $ADE$ , $DEFG$ , and $BCFG$ have the same perimeters, we can say the following:

$3x = x + 2(y-x) + y = y + 2(1-y) + 1$

Simplifying, we can find that

$3x = 3y-x = 3-y$

Since $3-y = 3x$ , $y = 3-3x$ .

After substitution, we find that $9-10x = 3x$ , and $x$ = $\frac{9}{13}$ .

Again substituting, we find $y$ = $\frac{12}{13}$ .

Therefore, $x+y$ = $\frac{21}{13}$ , which is $C$

@@ Line 1: / Line 1: @@
-Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y.
+Let <math>AD = x</math>, and <math>AG = y</math>. We want to find <math>DE + FG</math>, which is nothing but <math>x+y</math>.
-Based on the fact that ∆ADE, DEFG, and BCFG have the same perimeters, we can say the following:
+Based on the fact that <math>ADE</math>, <math>DEFG</math>, and <math>BCFG</math> have the same perimeters, we can say the following:
-x = x + 2(y-x) + y = y + 2(1-y) + 1
+<math>3x = x + 2(y-x) + y = y + 2(1-y) + 1</math>
 Simplifying, we can find that
-x = 3y-x = 3-y
+<math>3x = 3y-x = 3-y</math>
-Since 3-y = 3x, y = 3-3x.
+Since <math>3-y = 3x</math>, <math>y = 3-3x</math>.
-After substitution, we find that 9-9x-x = 3x, and x = 9/13.
+After substitution, we find that <math>9-10x = 3x</math>, and <math>x</math> = <math>\frac{9}{13}</math>.
-Again substituting, we find y = 12/13.
+Again substituting, we find <math>y</math> = <math>\frac{12}{13}</math>.
-Therefore, x+y = 21/13, which is C
+Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math>