Oblique Square Pyramids!

Geometry Level pending

A semicircle is inscribed in square base $ABCD$ as shown above and $BF$ is tangent to the semicircle at point $E$ .

Let the height of the oblique square pyramid $\overline{ET} = \overline{AB}$ and $A_{s}$ be the lateral surface area.

If $\dfrac{A_{s}}{A_{\square{ABCD}}} = \dfrac{\sqrt{\alpha} + \sqrt{\beta} + \sqrt{\gamma} + \sqrt{\lambda}}{\omega}$ ,

where $\alpha, \beta, \gamma, \lambda$ and $\omega$ are coprime positive integers, find $\alpha + \beta + \gamma + \lambda + \omega$ .

1 solution

Rocco Dalto
Mar 12, 2020

To find point $E(x,y)$ of square base $ABCD$ :

Let $a$ be the length of a side of the square $ABCD$ .

$(x - \dfrac{a}{2})^2 + y^2 = \dfrac{a^2}{2} \implies \dfrac{(2x - a)^2}{4} + y^2 = \dfrac{a^2}{4} \implies$

$x^2 - ax + y^2 = 0 \implies y = \sqrt{x(a - x)}$

$m_{BE} = \dfrac{\sqrt{x(a - x)} - a}{x} \implies m_{\perp} = m_{PE} =$ $\dfrac{2\sqrt{x(a - x)}}{2x - a} =$

$-\dfrac{x}{\sqrt{x(a - x)} - a} \implies$ $2ax - 2x^2 - 2a\sqrt{x(a - x)} = ax - 2x^2 \implies$

$x = 2\sqrt{x(a - x)} \implies x^2 = 4ax - 4x^2 \implies x(4a - 5x) = 0$ and $x \neq 0$

$\implies x = \dfrac{4a}{5} \implies y = \dfrac{2a}{5}$ $\implies E:(\dfrac{4a}{5}, \dfrac{2a}{5})$

Using $E:(\dfrac{4a}{5}, \dfrac{2a}{5})$ we obtain:

$UE = \dfrac{2a}{5} \implies RE = \dfrac{3a}{5}$ and $SE = \dfrac{4a}{5} \implies EQ = \dfrac{a}{5}$

$\implies S_{1} = \dfrac{\sqrt{29}}{5}a, S_{2} = \dfrac{\sqrt{26}}{5}a, S_{3} = \dfrac{\sqrt{34}}{5}a$ and $S_{4} = \dfrac{\sqrt{41}}{5}a \implies$

$A_{s} = a^2(\dfrac{\sqrt{29} + \sqrt{26} + \sqrt{34} + \sqrt{41}}{10}) \implies$

$\dfrac{A_{s}}{A_{\square{ABCD}}} = \dfrac{\sqrt{29} + \sqrt{26} + \sqrt{34} + \sqrt{41}}{10} =$

$\dfrac{\sqrt{\alpha} + \sqrt{\beta} + \sqrt{\gamma} + \sqrt{\lambda}}{\omega} \implies \alpha + \beta + \gamma + \lambda + \omega = \boxed{140}$ .