AMC 10 Weekly Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 23 Medium

As shown in the figure, circles M, N, and P are mutually externally tangent and also tangent internally to the equilateral triangle ABC. If a point is chosen at random inside equilateral triangle ABC, what is the probability that this point lies inside triangle MNP (the shaded region)?

  • A.

    \frac{\sqrt{3}-1}{2}

  • B.

    \frac{\sqrt{3}-1}{3}

  • C.

    \frac{2-\sqrt{3}}{2}

  • D.

    \frac{2-\sqrt{3}}{3}

  • E.

    \frac{\sqrt{2}-1}{2}

Answer:C

As shown in the figure, let the radius of one inscribed circle be r. Then

AH = BG = \sqrt{3}r,

so MN = GH = 2r, and AB = AH + BG + GH = 2(\sqrt{3}+1)r.

 

Since equilateral triangle MNP is similar to equilateral triangle ABC,

 

the probability that a randomly chosen point inside ABC lies in triangle MNP (the shaded region) is

P = \frac{A_{\triangle MNP}}{A_{\triangle ABC}} = \left(\frac{MN}{AB}\right)^{2} = \left(\frac{2r}{2(\sqrt{3}+1)r}\right)^{2} = \frac{2-\sqrt{3}}{2}.

 

Therefore, the answer is \text{C}.

Related Problems
Problem 21 AMC 10 Weekly Practice Round 3
Problem 22 AMC 10 Weekly Practice Round 3
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