<s>[INST] Maximising the Area of a Cyclic Quadrilateral In cyclic quadrilateral $ABCD$, $AB = AD$. If $AC = 6$ and $AB/BD = 3/5$, find the maximum possible value of $[ABCD]$.
(Source: SMT 2014)
If we let $AB=AD = 3x$ and $BD=5x$, from Ptolemy, we have $BC+CD=10$. From the triangle inequality, we can also get $BC+CD > BD \Rightarrow 10 > 5x \Rightarrow x < 2$. I'm stuck on where to go from here.
 [/INST] I've been exposed to a solution to this problem.
Since $AB=AD$, $\angle ABD = \angle ADB$. Furthermore, from the proprerties of a cyclic quadrilateral, $\angle ACB = \angle ADB$ and $\angle ACD = \angle ABD$. Therefore, $\angle ACB = \angle ADB = \angle ABD = \angle ACD$. Most importantly, $\angle ACB = \angle ACD$. Note that the area of $ABCD$ is equal to the area of $ABC$ added to the area of $ADC$. We use the alternate area formula $\frac{1}2 ab \sin C$, as follows: (Note that we let $\angle ACB = \angle ADB = \angle ABD = \angle ACD = \alpha$)
 \begin{align*}\\&[ABCD] \\&= [ABC]+[ADC] \\&=\frac{1}2(AC)(BC)\sin\alpha+\frac{1}2(AC)(DC)\sin\alpha \\&=\frac{1}2(AC)\sin\alpha(BC+CD)\end{align*} 
We are given $BC+CD = 10$ and $AC = 6$, so that $[ABCD] =\frac{1}2 (10)(6)(\sin \alpha) = 30 \sin \alpha$. Since $\alpha = \angle ADB$, we can compute the $\cos$ of $\alpha$ easily: drop the perpendicular from A to BD, and let X be the foot of that perpendicular. Note that $AXD$ is a right triangle with $DX = \frac{5}{2} x$, so $\cos \alpha = \frac{\frac{5}2}3 = \frac{5}6$. Therefore, since $\sin^2 \alpha + \cos^2 \alpha = 1$ and $\sin \alpha$ is positive, $\sin \alpha = \sqrt{11}{6}$, and the answer is $\boxed{5\sqrt{11}}$. The area is the same no matter what.

Thanks to the active members of the Art of Problem Solving Forums for posting this solution. Original link was here. The problem was lovely.
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