Trapazoids

Geometry Level 3

In the above isosceles trapezoids, $\overline{\rm AB}$ , $\overline{\rm CD}$ and $\overline{\rm EF}$ are parallel, $\angle{M_{2}M_{3}F}$ and $\angle{M_{2}M_{1}B}$ are right angles and $M_{1},M_{2},$ and $M_{3}$ are midpoints of $\overline{\rm AB},\overline{\rm CD}$ , and $\overline{\rm EF}$ respectively. The circle inscribed in isosceles trapezoid $ECDF$ is tangent to the trapezoid at points $M_{2},M_{3},Q$ and $P$ . The height of isosceles trapezoid $ACDB$ is $\overline{\rm M_{1}M_{2}} = 4$ , the lower base $\overline{\rm AB} = 2(1 + \sqrt{3})$ , and the upper base to both isosceles trapezoids $ECDF$ and $ACDB$ is $CD = 2$ .

Using the information in the above diagram, find the radius $R$ of the circle inscribed in isosceles trapezoid $ECDF$ to six decimal places.

The answer is 1.522737.

1 solution

Rocco Dalto
Feb 28, 2018

In the diagram above $\angle{DHF}$ is a right angle.

Using the pythagorean theorem on right $\triangle{DHF} \implies 4R^2 + (w - 1)^2 = (w + 1)^2$ $\implies R = \sqrt{w}$ and $\triangle{DGB} \sim \triangle{DHF}$ $\implies \dfrac{\sqrt{3}}{w - 1} = \dfrac{4}{2R} \implies R = \dfrac{2(w - 1)}{\sqrt{3}} \implies$

$\dfrac{2(w - 1)}{\sqrt{3}} = \sqrt{w} \implies 4w^2 - 8w + 4 = 3w \implies 4w^2 - 11w + 4 = 0 \implies$

$w = \dfrac{11 \pm \sqrt{57}}{8}$ .

$0 < w = \dfrac{11 - \sqrt{57}}{8} < \dfrac{1}{2} \implies R = \dfrac{2(w - 1)}{\sqrt{3}} < 0$ and $R = \sqrt{w} > 0 \therefore$ drop $w = \dfrac{11 - \sqrt{57}}{8}$

Choosing $w = \dfrac{11 + \sqrt{57}}{8} > 1 \implies R = \dfrac{2(w - 1)}{\sqrt{3}} = \sqrt{w} \approx \boxed{1.522737}$

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/

Jake Tricole - 1 year, 12 months ago

Trapazoids

The answer is 1.522737.

1 solution

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