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A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot be greater than sqrt2. Find a quadrilateral of this type for which s2 is approximately sqrt3and show that s2 is always less than sqrt3. Find a quadrilateral of this type for which s3 is approximately 2 and show that s2 is always less than 2. Find a quadrilateral of this type for which s4=2 and show that s4 cannot be greater than 2.

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Diagonals for Area

Prove that the area of a quadrilateral is given by half the product of the lengths of the diagonals multiplied by the sine of the angle between the diagonals.

Cyclic Quad Jigsaw

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

We had two very clear solutions to this problem - well done to James, from Poole Grammar School, and Dylan, who did not give his school. James's solution is shown here.

Cyclic Quadrilateral diagram
All five of the small quadrilaterals in the above shape are cyclic, and by using the fact that opposite angles in a cyclic quadrilateral add up to $180 ^{\circ}$ we can prove that the large quadrilateral made up of the five smaller ones is also cyclic.

Opposite angles in each cyclic quadrilateral add up to $180^{\circ}$, so we can write expressions for angles $a$, $c$, and $e$.

$$a = 180^{\circ}- b$$ $$c = 180^{\circ}- d$$ $$e = 180^{\circ}- f$$

Angles $a$, $c$, and $e$ are all round the same point therefore: $$360^{\circ} = a + c + e$$ Substituting in the expressions for angles c and e in we get: $$360^{\circ} = a + (180^{\circ} - d) + (180^{\circ} - f)$$

Simplifying this we get: $$360^{\circ} = a + 360^{\circ} - d - f$$ $$ a = d + f$$

Now we follow exactly the same working with the other side of the quadrilateral which gives us the equation $$b = h + j$$

We also know that $a + b = 180^{\circ}$ because they are opposite angles in a cyclic quadrilateral. Substituting in the expressions for $a$ and $b$ gives us $$d + f + h + j = 180^{\circ}$$
So $$(d + h) = 180^{\circ} - (f + j)$$

Looking back at the original diagram we can see that $(d + h)$ and $(f +j)$ are opposite angles in the quadrilateral and because $(d + h) = 180^{\circ} - (f + j)$ the quadrilateral must be cyclic.