AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 11 Medium

Triangle ABC is equilateral with AB=1. Points E and G are on \overline{AC} and points D and F are on \overline{AB} such that both \overline{DE} and \overline{FG} are parallel to \overline{BC}. Furthermore, triangle ADE and trapezoids DFGE and FBCG all have the same perimeter. What is DE+FG?

  • A.

    1

  • B.

    \frac{3}{2}

  • C.

    \frac{21}{13}

  • D.

    \frac{13}{8}

  • E.

    \frac{5}{3}

Answer:C

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.

Related Problems
Problem 9 AMC 10 Daily Practice Round 3
Problem 10 AMC 10 Daily Practice Round 3
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