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ABCD is an isosceles trapezoid.

$CAB = \alpha,\quad CAD = \beta,\quad AC = m$

a) Find $AB$ and $DC$.
b) $\beta = 2\alpha$, $\frac{AB}{DC}$ = $\frac{1}{2}$, Find $a$ without using $β$ or $m$.
c) Is it possible to inscribe a circle in $ABCD$? Why?

Answers:
a) $AB = \frac{m\sin(2\alpha+\beta)}{\sin(\alpha+\beta)}$, $DC = \frac{m\sin(\beta)}{\sin(\alpha+\beta)}$

b) $\alpha = 37.76^\circ$

c) No.

I've managed to solve a, but I'm having trouble solving b.

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Part (b) follows directly from part (a). We have $\dfrac{AB}{DC}=\dfrac{1}{2}$. Using the expressions you got for $AB$ and $CD$, putting $\beta=2\alpha$, and simplifying a bit, we find that $\frac{1}{2}=\frac{AB}{CD}=\frac{\sin 4\alpha}{\sin 2\alpha}$ (there is a fair amount of cancellation.)

Now use the double-angle identity $\sin 4\alpha=2(\sin 2\alpha)(\cos 2\alpha)$. We find that $\cos 2\alpha =\frac{1}{4}.$ The rest is a job for the calculator, use the $\cos^{-1}$ button.

If we feel like it we can get an explicit expression for $\cos \alpha$ or $\sin \alpha$, from the fact that $\cos 2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$, but that is not necessary.

As to part (c), one cannot inscribe a circle in any trapezoid that satisfies the conditions of (b). First draw an isosceles trapezoid with an inscribed circle. Suppose that the two tangents to a circle from an external point $P$, meet the circle at $M$ and $N$. Then $PM=PN$. So in any isosceles trapezoid with an inscribed circle, each "slant" side has length equal to half the sum of the two parallel sides.

In our case, we can take $AB=2$ and $CD=4$. So if our trapezoid had an inscribed circle, then each slant side would have length $3$. But then the height of the trapezoid would be $\sqrt{3^2-1^2}$, that is, $2\sqrt{2}$. That makes $\tan\alpha =\frac{2\sqrt{2}}{3}$. But then $\angle \alpha\approx 43.3$ degrees, which is not what we found in part (b).

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    Got it, thanks again!2012-04-13