Monday, February 17, 2014

2014 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of an equilateral triangle

#1: P is the interior of equilateral triangle ABC, such that perpendicular segments from P to each of the sides of triangle ABC measure 2 inches, 9 inches and 13 inches. Find the number of square inches in the area of triangle ABC, and express your answer in simplest radical form.

#2: AMC 2007-B: Point P is inside equilateral  ABC. Points Q, R, and S are the feet of the
perpendiculars from P to AB, BC, and CA, respectively. Given that PQ = 1,PR = 2, and PS = 3, what is AB ?

This is an equilateral triangle. If the side is "S", the length of the in-radius would be\(\dfrac {\sqrt {3}} {6}\) of  S (or \(\dfrac {1} {3}\)of the height) and the length of the circum-radius would be\(\dfrac {\sqrt {3}} {3}\) of  S (or \(\dfrac {2} {3}\)of the height).

You can use 30-60-90 degree special right triangle angle ratio to get the length of each side as well as the height.

Solution I: Let the side be "s" and break the triangle into three smaller triangles.

\(\dfrac {9+13+2} {2}\)= 12s (base times height divided by 2)= \(\dfrac {\sqrt {3}} {4}\times s^{2}\)

s = 16 3 

Area of the triangle = 192 3


Solution II: Let the side be "s" and the height of the equilateral triangle be "h"
24* s (by adding 9, 2 and 13 since they are the height of each smaller triangle)  = s*h 
(Omit the divided by 2 part on either side since it cancels each other out.)
h = 24
Using 30-60-90 degree angle ratio, you get \(\dfrac {1} {2}\) s = 8 3  so  s = 16 3 
Area of the equilateral triangle = \(\dfrac {24\times 16\sqrt {3}} {2}\) = \(192\sqrt {3}\) 

#2: This one is similar to #1, the answer is 4 3 

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