2007 AMC 10B Problem 4

Below is the professionally curated solution for Problem 4 of the 2007 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10B solutions, or check the answer key.

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Concepts:circumcircle, circumcenter, and circumradiusisosceles triangleangle chasing

Difficulty rating: 1030

4.

The point OO is the center of the circle circumscribed about ABC,\triangle ABC, with BOC=120\angle BOC = 120^\circ and AOB=140,\angle AOB = 140^\circ, as shown. What is the degree measure of ABC?\angle ABC?

3535

4040

4545

5050

6060

Solution:

Since OA=OB=OC,OA=OB=OC, triangles AOB,BOC,AOB, BOC, and COACOA are isosceles. The base angles give ABO=1801402=20\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ and OBC=1801202=30.\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.

Therefore ABC=20+30=50.\angle ABC=20^\circ+30^\circ=50^\circ.

Thus, the correct answer is D.

Problem 4 in Other Years