Circles and Squares!!

Level pending

In square $ABCD$ , one of the vertices of square $MPCN$ touches $\overline{EF}$ at $P$ and $\overline{EF}$ is tangent to the above circle at $P$ and the radius of the circle is half the side of the square $MPCN$ .

Let $A_{T}$ be the area of the green shaded regions.

If $\dfrac{A_{T}}{A_{ABCD}} = \dfrac{a + b\sqrt{c}}{d}$ , where $a,b,c$ and $d$ are coprime positive integers, find $a + b + c + d$ .

The answer is 370.

1 solution

Rocco Dalto
Dec 10, 2020

Using the above diagram $\overline{CB} = \sqrt{2}a = \sqrt{2}x + \dfrac{x}{2} + \dfrac{x}{\sqrt{2}}$ $\implies (6 + \sqrt{2})x = 4a \implies x = \dfrac{4a}{6 + \sqrt{2}}$ $= \dfrac{4(6 - \sqrt{2})}{34}a = \dfrac{12 - 2\sqrt{2}}{17}a$

$PB = \dfrac{x}{2} + \dfrac{x}{\sqrt{2}} = \dfrac{2 + \sqrt{2}}{2\sqrt{2}}x = (\dfrac{2 + \sqrt{2}}{2\sqrt{2}})(\dfrac{12 - 2\sqrt{2}}{17})a$ $= \dfrac{(\sqrt{2} + 1)(6 - \sqrt{2})}{17}a = \dfrac{5\sqrt{2} + 4}{17}a$

$\implies EB = \sqrt{2}PB = \sqrt{2}(\dfrac{5\sqrt{2} + 4}{17})a = \dfrac{10 + 4\sqrt{2}}{17}a$

$\implies A_{T} = a^2 -(x^2 + A_{\triangle{EBF}}) = a^2 - (x^2 + \dfrac{1}{2}EB^2) =$

$(1 - (\dfrac{(12 - 2\sqrt{2})^2}{289}+ 2\dfrac{(5 + 2\sqrt{2})^2}{289}))a^2$ $= \dfrac{a^2}{289}(289 - 218 + 8\sqrt{2}) =$

$\dfrac{71 + 8\sqrt{2}}{289}a^2 \implies \dfrac{A_{T}}{A_{ABCD}} = \dfrac{71 + 8\sqrt{2}}{289} =$ $\dfrac{a + b\sqrt{c}}{d} \implies a + b + c + d = \boxed{370}$ .

Circles and Squares!!

The answer is 370.

1 solution

0 pending reports