2016 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2016 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:special right trianglePythagorean Theorem3D geometry

Difficulty rating: 1630

13.

Alice and Bob live 1010 miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is 3030^\circ from Alice's position and 6060^\circ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?

3.53.5

44

4.54.5

55

5.55.5

Solution:

Let the airplane be at C,C, directly above point DD on the ground at altitude h.h. Triangles ACDACD and BCDBCD are 3030-6060-9090 right triangles, so AD=3hAD=\sqrt3\,h and BD=h3.BD=\dfrac{h}{\sqrt3}. Since Alice looks north and Bob looks west, ADB=90,\angle ADB=90^\circ, so AD2+BD2=AB2=100.AD^2+BD^2=AB^2=100. Then 3h2+h23=10h23=100,3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100, giving h=305.48,h=\sqrt{30}\approx5.48, closest to 5.5.5.5.

Thus, the correct answer is E.

Problem 13 in Other Years