2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 19 Hard

Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S? (2019 AMC 10B Problem, Question#19)

  • A.

    98

  • B.

    100

  • C.

    117

  • D.

    119

  • E.

    121

Answer:C

Divide the circle into four parts: the top semicircle \text {(A)}; the bottom sector \text {(B)}, whose arc angle is 120^\circ because the large circle's radius is 2 and the short length(the radius of the smaller semicircles) is 1, giving a 30^\circ—60^\circ—90^\circ triangle; the triangle formed by the radii of A and the chord \text {(C)}, and the four parts which are the corners of a circle inscribed in a square \text {(D)}, Then the area is A+B-C+D (in B-C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the area of the bottom shaded region).

The area of A is \frac12\pi ·2^2=2\pi,

The area of B is \frac{120^{\circ}}{360^{\circ}}\pi\cdot 2^{2}= \frac{4 \pi}{3}.

For the ara of C, th. radius of 2, and he distance of 1 (the smaller semicicles' radius ) to BC．

creates two 30^\circ—60^\circ—90^\circ triangles, so C's area is 2\cdot \frac 12\cdot 1\cdot \sqrt{3}=\sqrt{3},

The area of D is {4}\cdot 1- \frac{1}{4}\pi \cdot 2^{2}=4-\pi,

Hence, finding A+B-C+D. the desired areas is \frac{7 \pi}{3}- \sqrt{3}+4, so the answer is 7+3+3+4=\text {(E)}17.

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