Limit of Ptolemy's theorem

Geometry Level 3

Is it possible to construct a cyclic quadrilateral $ABCD$ such that $AB = 17$ , $BC = 18$ , $CD = 29$ , $DA = 25$ , $AC = 41$ and $BD =23$ ?

No Cannot be determined Yes

1 solution

Tommy Li
Sep 20, 2017

Relevant wiki: Ptolemy's Theorem

Ptolemy's theorem state that : If a quadrilateral is inscribable in a circle then the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

$17\times 29 +18 \times 25 = 41 \times 23 = 943$

$AB \times CD +BC \times DA = AC \times BD$

However , consider $\Delta ABC$ and think about the triangle inequality :

$17 +18 < 41 \Rightarrow AB + BC < CA$

The quadrilateral $ABCD$ doesn't exist and we can't use Ptolemy's theorem to say yes to the problem

Great! That is a necessary, but not sufficient condition for a cyclic quadrilateral.

Any thoughts on what a sufficient condition would be?

Calvin Lin Staff - 3 years, 8 months ago

It won't work for $AC=41, BD=23$ either, even though this condition is not an obstacle anymore. For a given set of side lengths, diagonals cannot be just arbitrarily picked to match the formula. The diagonal lengths can be determined by using the law of cosines:

In this case, $AC=\sqrt{920704651} / 1031, BD=\sqrt{920704651}/947$

Maria Kozlowska - 3 years, 7 months ago

Right. As it turns out, the lengths of the diagonals are uniquely determined.

There is a "second Ptolemy theorem" which states that $\frac{ AC} { BD} = \frac{ AB \times AD + CB \times CD } { AB \times BC + CD \times DA }$ . (This could also be proven by the law of cosines). Hence, the cyclic quadrilaterial is uniquely determined by the 4 sides on the perimeter.

Calvin Lin Staff - 3 years, 7 months ago

Limit of Ptolemy's theorem

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