2007 AMC 12A Problem 23

Attempt Problem 23 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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23.

Square ABCDABCD has area 36,36, and ABAB is parallel to the xx-axis. Vertices A,A, B,B, and CC are on the graphs of y=logax,y=\log_a x, y=2logax,y=2\log_a x, and y=3logax,y=3\log_a x, respectively. What is a?a?

36\sqrt[6]{3}

3\sqrt3

63\sqrt[3]{6}

6\sqrt6

66

Answer: A
Concepts:logarithmcoordinate geometrysquare (geometry)
Difficulty rating: 1990
Solution:

Let A=(p,logap)A=(p,\log_a p) and B=(q,2logaq).B=(q,2\log_a q). Since ABAB is horizontal, logap=2logaq=logaq2,\log_a p=2\log_a q=\log_a q^2, so p=q2.p=q^2.

The side length is 6=pq=q2q,6=|p-q|=|q^2-q|, whose only positive solution is q=3.q=3.

Since C=(q,3logaq),C=(q,3\log_a q), the vertical side gives BC=6=logaq=loga3.BC=6=\log_a q=\log_a 3. Thus a6=3,a^6=3, so a=36.a=\sqrt[6]{3}.

Thus, the correct answer is A.

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