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Geometry Difficulty 4.5 AIME Find the answer Canada

In the diagram, point $C$ is on side $BD$ of quadrilateral $ABDE$ . Also, $AB$ and $ED$ are perpendicular to $BD$ , $\angle ACB = 60\degree$ , $\angle CAE = 45\degree$ , and $\angle AEC = 45\degree$ .

If $AB = \sqrt{3}$ , what is the perimeter of quadrilateral $ABDE\,$ ?

Pick one

Solution

We note first that $\triangle ACB$ has a right angle and a $60\degree$ angle and so it is a $30\degree$ - $\,60\degree$ - $\,90\degree$ triangle. Since $AB = \sqrt{3}$ , then using the known ratios of side lengths, we can see that $BC = 1$ and $AC = 2$ .

Next, we note that $\triangle ACE$ has two $45\degree$ angles and so is an isosceles right-angled triangle. This means that $CE = AC = 2$ and $\angle ACE = 90\degree$ . Also, $AE = $\sqrt{2}AC = 2\sqrt{2}$ $. Further, since$ $\angle$ BCD $is a straight angle, then$ $\angle ECD = 180\degree - \angle ACB - \angle ACE = 180\degree - 60\degree - 90\degree = 30\degree$ $Since$ $\triangle$ CED $has a$ $30\degree$ $angle and a right-angle, it is also a$ $30\degree$ $-$ $\,60\degree$ $-$ $\,90\degree$ triangle. Using the known ratios of sides, since

Figure for this problem CE =
2 $, we have$ DE = 1 $and$ CD = $\sqrt{3}$ $. Therefore, the perimeter of$ ABDE$ is
$AB + BC + CD + DE + AE = \sqrt{3} + 1 + \sqrt{3} + 1 + 2\sqrt{2} = 2 + 2\sqrt{2} + 2\sqrt{3}$

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