Dynamic Geometry: P104

Geometry Level 4

The diagram shows a blue equilateral triangle with side length one. Three orange rectangles are growing and shrinking at the same rate, their commom points creates a purple hexagone. When the ratio of the hexagone's area to the area of one orange rectangle is equal to $\dfrac{1}{4}$ , the product of their area can be expressed as $\dfrac{p}{q}$ , where $p$ and $q$ are coprime positive integers. Find $p+q$ .

The answer is 628.

1 solution

Chew-Seong Cheong
Apr 1, 2021

Let the width of the rectangle be $a$ , its height $h$ , and the side length of the hexagon be $b$ . Then $h = \dfrac {\sqrt 3}2 - \dfrac {\sqrt 3}2 a$ , the area of the rectangle $A_r = \dfrac {\sqrt 3}2 (1-a)a$ , $b = \dfrac a2 \cdot \dfrac 2{\sqrt 3} = \dfrac a{\sqrt 3}$ , and the area of the hexagon $A_h = 6 \cdot \dfrac 12 \cdot b^2 \sin 60^\circ = \dfrac {\sqrt 3}2a^2$ . When

$\begin{aligned} \frac {A_h}{A_r} & = \frac 14 \\ 4A_h & = A_r \\ 4 \cdot \frac {\sqrt 3}2a^2 & = \frac {\sqrt 3}2(1-a)a \\ 4a & = 1 -a \\ \implies a & = \frac 15 \\ \implies A_hA_r & = 4A_h^2 = 4 \left(\frac {\sqrt 3}2a^2 \right)^2 = \frac 3{625} \end{aligned}$

Therefore $p+q = 3+625 = \boxed{628}$ .

Let the side of the hexagon be considered x, and then chase lengths through 30-60-90 and 120-30-30 triangles, and come up with this answer. This is how I solved it, without trig, though there was a bit of algebra.

Sanchit Sharma - 2 months, 1 week ago

Dynamic Geometry: P104

The answer is 628.

1 solution

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