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In the figure below, ABC is equilateral. P is a point inside ABC such that BPC is a right isosceles triangle. What is ABP in degrees?

https://latex.artofproblemsolving.com/7/1/d/71daedfa74cc903aeb2620c72d5d2ec34dbfcfd8.png

 Sep 24, 2019
 #1
avatar+142 
-1

 

Image

In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270



Look at the diagram below:

Image

Each of the three red central angles is 120° (360°/3 = 120°). Thus in order point B to be in the position where point A is, the triangle should be rotated clockwise by 120° + 120° = 240°.

Answer: D.

 Sep 24, 2019
 #2
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+1

Since ABC is an equilateral triangle, that means that each of the 3 angles, A, B, C = 60 degrees.
Also, since BPC is a right isosceles triangle, that means the 2 angles, PBC =PCB =45 degrees each, since angle BPC = 90 degrees.
Therefore, angle ABP = 60 - 45 = 15 degrees.

 Sep 24, 2019

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