Welcome to our community

Be a part of something great, join today!

A quadrilateral inscribed in a semicircle

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,725

trig.png
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 $x^3 = (a^2 + b^2 + c^2)x + 2abc.$
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753

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 $x^3 = (a^2 + b^2 + c^2)x + 2abc.$

Thank you Opalg for participating!

My solution:

From the Ptolemy Theorem, we have the product of diagonals equals the sum of the products of the opposite sides for any cyclic quadrilateral.

A quadrilateral inscribed in a semicircle.JPG
So, in our case, we have the following identity from the Ptolemy Theorem:

$PR \cdot QS=ac+bx$---(1)

In order to get rid of the variables $PR$ and $QS$, we have to relate them to the variables $a, b, c, x$ and this can be done by observing there are two right-angle triangles exist in the diagram since PS is the diameter of the cicle, so by applying the Pythagoras' Theorem to each of these triangles we get:

$PR^2=x^2-c^2$ and $QS^2=x^2-a^2$

Hence, raise the equation (1) to the second power and replace the two equations above into it gives

$PR^2 \cdot QS^2=a^2c^2+b^2x^2+2abcx$

$(x^2-c^2)(x^2-a^2)=a^2c^2+b^2x^2+2abcx$

$x^4-a^2x^2-c^2x^2+\cancel{a^2c^2}=\cancel{a^2c^2}+b^2x^2+2abcx$

$x^4-a^2x^2-c^2x^2=b^2x^2+2abcx$

$x^4-(a^2+b^2+c^2)x^2-2abcx=0$

$\therefore x^3-(a^2+b^2+c^2)x-2abc=0$ since $x\ne 0$ and we're done.