Not trigo, please

Geometry Level 2

Each side of the four squares shown in the figure above is of length 1 unit. Points $A, B, C$ and $D$ are vertices of the squares. $AC$ and $BD$ are straight lines that intersect at point $E$ . Suppose $\angle AEB = x^{\circ}$ . Find $x$ .

The answer is 45.

2 solutions

Chris Lewis
May 13, 2019

$DF$ is parallel to $AC$ , so the two grey shaded angles are both $x$ .

Clearly $BFD$ is a right-angled, isosceles triangle, so $x=\boxed{45^{\circ}}$

David Vreken
May 13, 2019

The angle $x$ between two lines with slopes $m_1$ and $m_2$ is $\tan x = \frac{m_1 - m_2}{1 + m_1m_2}$ .

In this case, $AC$ has a slope of $m_1 = 2$ and $BD$ has a slope of $m_2 = \frac{1}{3}$ , so that $\tan x = \frac{2 - \frac{1}{3}}{1 + 2 \cdot \frac{1}{3}}$ or $\tan x = 1$ , which means $x = \boxed{45°}$ .

@David Vreken from the diagram drawn by @Chris Lewis, we may get the following equations: $\tan^{-1}2-\tan^{-1}\frac{1}{3} = \frac{\pi}{4}$ $\tan^{-1}3-\tan^{-1}\frac{1}{2} = \frac{\pi}{4}$ $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3} = \frac{\pi}{4}$

Watch this for the animation.

Chan Lye Lee - 2 years ago

I just realized that the title of the problem meant that we were supposed to it without trigonometry. Sorry about that!

David Vreken - 2 years ago

No worries at all.

Chan Lye Lee - 2 years ago

Not trigo, please

The answer is 45.

2 solutions

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