Which Triangle is Larger?

Geometry Level 4

A E B F D AEBFD is a regular pentagon of side length 1. A B C ABC is an equilateral triangle with a side A B AB congruent to the pentagon's diagonal. Which triangle, red or blue, has the larger area? For extra credit, provide closed-form representations of the areas of the red and blue triangles.

Red Triangle Blue Triangle Neither Triangle

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1 solution

Charley Shi
Apr 15, 2021

All angles are in degrees. Line segment E H \overline{EH} divides A B E \triangle {ABE} into two congruent right triangles. This gives the length of A B \overline{AB} as A B = 2 1 cos ( 36 ) = 2 cos ( 36 ) \overline{AB} = 2\cdot 1\cdot \cos(36) = 2\cos(36) Using the sine rule on B F G \triangle BFG , we can calculate the sides F G FG and B G BG . 1 sin ( 60 ) = F G sin ( 12 ) = B G sin ( 108 ) \frac{1}{\sin(60)} = \frac{\overline{FG}}{\sin(12)} = \frac{\overline{BG}}{\sin(108)} F G = sin ( 12 ) sin ( 60 ) B G = sin ( 108 ) sin ( 60 ) \overline{FG} = \frac{\sin(12)}{\sin(60)} \quad \overline{BG} = \frac{\sin(108)}{\sin(60)} Since A B C \triangle ABC is equilateral, B G + G C = A B \overline{BG} + \overline{GC} = \overline{AB} . Consequently, G C = A B B G = 2 cos ( 36 ) sin ( 108 ) sin ( 60 ) \overline{GC} = \overline{AB} - \overline{BG} = 2\cos(36) - \frac{\sin(108)}{\sin(60)} Use the area formula A = 1 2 a b sin ( C ) A = \frac{1}{2}ab \sin(C) on both the blue and red triangles. A ( blue ) = 1 2 1 sin ( 12 ) sin ( 60 ) sin ( 108 ) 0.114 A(\textrm{blue}) = \frac{1}{2}\cdot 1 \cdot \frac{\sin(12)}{\sin(60)} \cdot \sin(108) \approx 0.114 A ( red ) = 1 2 ( 2 cos ( 36 ) sin ( 108 ) sin ( 60 ) ) 2 sin ( 60 ) 0.117 A(\text{red}) = \frac{1}{2} \cdot \left(2\cos(36) - \frac{\sin(108)}{\sin(60)}\right)^2\cdot \sin(60) \approx 0.117 Therefore, the red triangle has a larger area than the blue triangle.

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