A geometry problem by A Former Brilliant Member

Geometry Level 1

Circle O O is inscribed in A B C , \triangle ABC, as shown in the figure above. The points of tangency are at D , E , D,E, and F F . Given that A D = 2 AD=2 and D C = 3 DC=3 , find the area of A B C \triangle ABC .


The answer is 6.

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1 solution

Let E B = B F = x EB=BF=x . By Pythagorean theorem , we have

25 = ( x + 2 ) 2 + ( x + 3 ) 2 25=(x+2)^2+(x+3)^2

25 = x 2 + 4 x + 4 + x 2 + 6 x + 9 25=x^2+4x+4+x^2+6x+9

Simplifying further, we get

x 2 + 5 x 6 = 0 x^2+5x-6=0

By factoring, we get

x = 1 x=1

Finally, the area of A B C \triangle ABC is 1 2 ( B C ) ( A B ) = 1 2 ( 4 ) ( 3 ) = \dfrac{1}{2}(BC)(AB)=\dfrac{1}{2}(4)(3)= 6 \boxed{6}

What a nice problem. Waiting for another.

christian reivan - 3 years ago

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