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anemone
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Let $PQRS$ be a quadrilateral inscribed in a semicircle with diameter $PS=x$. If $PQ=a$, $QR=b$, $RS=c$, then prove that $x^3-(a^2+b^2+c^2)x-2abc=0$.
Opalg said:[sp]
Denote the angles at $O$ as $2\alpha$, $2\beta$, $2\gamma$, as in the diagram. Note that $\sin\alpha = \dfrac{a/2}{x/2} = \dfrac ax$, and similarly $\sin\beta = \dfrac bx$, $\sin\gamma = \dfrac cx.$ Also, $\alpha + \beta + \gamma = \pi/2$, so that $\sin\alpha = \cos(\beta + \gamma) = \cos\beta\cos\gamma - \sin\beta \sin\gamma$. Hence $\cos\beta\cos\gamma = \sin\alpha +\sin\beta \sin\gamma$, and similarly $\cos\alpha\cos\gamma = \sin\beta +\sin\alpha \sin\gamma$. Then $$\begin{aligned}1 = \sin(\alpha+\beta+\gamma) &= \sin(\alpha+\beta)\cos\gamma + \cos(\alpha+\beta)\sin\gamma \\ &= \sin\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\cos\gamma + \sin^2\gamma \\ &= \sin\alpha(\sin\alpha + \sin\beta\sin\gamma) + \sin\beta(\sin\beta + \sin\alpha\sin\gamma) + \sin^2\gamma \\ &= \sin^2\alpha + \sin^2\beta + \sin^2\gamma + 2\sin\alpha\sin\beta\sin\gamma. \end{aligned}$$ Now all you have to do is to substitute the values $\sin\alpha = \dfrac ax$, $\sin\beta = \dfrac bx$, $\sin\gamma = \dfrac cx$ to get $1 = \dfrac{a^2}{x^2} + \dfrac{b^2}{x^2} + \dfrac{c^2}{x^2} + 2\dfrac{abc}{x^3}$, from which $x^3 = (a^2 + b^2 + c^2)x + 2abc.$[/sp]
A quadrilateral inscribed in a semicircle is a four-sided shape that is contained within a semicircle, meaning that all four of its vertices touch the semicircle's circumference.
The total angle sum of a quadrilateral inscribed in a semicircle is always 180 degrees. This is because the two angles opposite each other in a quadrilateral inscribed in a semicircle are always supplementary (add up to 180 degrees).
The angles in a quadrilateral inscribed in a semicircle have a special relationship known as "opposite angles are supplementary." This means that the two angles opposite each other in the quadrilateral add up to 180 degrees.
No, a quadrilateral is not always inscribed in a semicircle. In order for a quadrilateral to be inscribed in a semicircle, all four of its vertices must touch the semicircle's circumference.
The area of a quadrilateral inscribed in a semicircle can be found by dividing the semicircle into two equal halves and finding the area of each half (a semicircle is half of a circle). Then, we can use the formula for the area of a triangle (A = 1/2 * base * height) to find the area of each triangle. Finally, add the areas of the two triangles together to find the total area of the quadrilateral inscribed in the semicircle.