MHB A quadrilateral inscribed in a semicircle

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Inscribed
AI Thread Summary
In the discussion, a proof is presented for the equation relating the sides of a quadrilateral inscribed in a semicircle, specifically showing that \(x^3 - (a^2 + b^2 + c^2)x - 2abc = 0\), where \(x\) is the diameter and \(a\), \(b\), and \(c\) are the lengths of the sides. The proof utilizes trigonometric identities, particularly the sine and cosine functions of angles formed at the center of the semicircle. By expressing the sine values in terms of the sides and substituting them into the equation, the relationship is established. The angles at the center sum to \(\pi/2\), leading to a simplification that confirms the derived equation. This mathematical relationship highlights the geometric properties of quadrilaterals inscribed in semicircles.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $PQRS$ be a quadrilateral inscribed in a semicircle with diameter $PS=x$. If $PQ=a$, $QR=b$, $RS=c$, then prove that $x^3-(a^2+b^2+c^2)x-2abc=0$.
 
Mathematics news on Phys.org
[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]
 

Attachments

  • trig.png
    trig.png
    6.2 KB · Views: 143
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]

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.

View attachment 1752
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.
 

Attachments

  • A quadrilateral inscribed in a semicircle.JPG
    A quadrilateral inscribed in a semicircle.JPG
    19.2 KB · Views: 136
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top