5 Circles and a Trapezoid

Geometry Level 4

The figure to the right shows 5 unit circles that are tangent to each other and the edges of the trapezoid. If the yellow area can be expressed as a + b c d π a+\frac{b}{\sqrt{c}}-d\pi (such that a , b , c , and d a,~b,~c,~\text{and}~d are integers and c c is square-free), then find a + b + c d \frac{a+b+c}{d} .


Note: I did not create this problem; I simply solved it and decided to post it on here. Credit goes to my math teacher.


The answer is 6.0.

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1 solution

The area of the yellow region A yellow A_{\color{#CEBB00}\text{yellow}} is given by the area of the trapezoid A trapezoid A_{\text{trapezoid}} minus the area of the 5 circles A circle A_{\text{circle}} .

From the figure, we note that the length of the base of the trapezoid a b = 4 + 2 a = 4 + 2 tan 3 0 = 4 + 2 3 a_b = 4+2a = 4+\dfrac 2{\tan 30^\circ} = 4+2\sqrt 3 . The length of its top a t = 2 + 2 b = 2 + 2 tan 3 0 = 2 + 2 3 a_t = 2+2b = 2 + 2\tan 30^\circ = 2 + \dfrac 2{\sqrt 3} . The height of the trapezoid h = 2 + c = 2 + 2 sin 6 0 = 2 + 3 h = 2+c = 2+2 \sin 60^\circ = 2+\sqrt 3 .

Therefore, we have:

A yellow = A trapezoid 5 A circle = ( a b + a t ) h 2 5 π 1 2 = 1 2 ( 4 + 2 3 + 2 + 2 3 ) ( 2 + 3 ) 5 π = 10 + 17 3 5 π \begin{aligned} A_{\color{#CEBB00}\text{yellow}} & = A_{\text{trapezoid}} - 5A_{\text{circle}} \\ & = \frac {(a_b+a_t)h}2 - 5\pi 1^2 \\ & = \frac 12\left(4+2\sqrt 3+2+\frac 2{\sqrt 3}\right)(2+\sqrt 3) - 5\pi \\ & = 10 + \frac {17}{\sqrt 3} - 5\pi \end{aligned}

Then a + b + c d = 10 + 17 + 3 5 = 6 \dfrac {a+b+c}d = \dfrac {10+17+3}5 = \boxed 6 .

How do you know the angle is 60°??

Valentin Duringer - 1 year, 3 months ago

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Place the sixth circle with the same radius on top and extend the two slanting sides. You will find that the two slanting sides will meet a equilateral triangle is formed. Therefore the base angles of the trapezium are 6 0 60^\circ .

Chew-Seong Cheong - 1 year, 3 months ago

thank you for your reply

Valentin Duringer - 1 year, 3 months ago

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