Problem 642

Solution:

Since $AB^{2} + AD^{2} = BD^{2}$ and $BC^{2} + CD^{2} = BD^{2}$ the quadrilateral $ABCD$ can be inscribed in a circle of diameter $BD$. Let $DH$ be the altitude of triangle $DAC$. Then triangle $AHD$ is similar to triangle $BCD$ because they are both right triangles and $\widehat{CBD} = \widehat{CAD} = \widehat{HAD}$, since they are inscribed angles subtending $CD$. Therefore $BD : BC = AD : AH$, that is $25 : 20 = 7 : AH$ from which we obtain that the length of $AH$ is $28/5$. Similarly triangle $CDH$ is similar to triangle $BDA$, from which $BD : AB = CD : HC$, that is $25 : 24 = 15 : HC$, which gives $HC = 72/5$. The length of $AC$ therefore turns out to be $AH + HC = (28 + 72)/5 = 20$.

One can observe that $25^{2} = 20^{2} + 15^{2}$, that is $(15,20,25)$ is a Pythagorean triple $(5 \cdot (3,4,5))$, hence the angle $\widehat{BCD}$ is right. We also have $25^{2} = 24^{2} + 7^{2}$: also $(7,24,25)$ is a Pythagorean triple and the angle $\widehat{DAB}$ is right.

Let $AK$ and $CL$ be the altitudes of the right triangles $ABD$ and $BCD$ relative to $BD$; we can obtain their lengths from the areas of the two triangles: $AK = (AD \cdot AB)/BD = (24 \cdot 7)/25$, while $CL = (BC \cdot CD)/BD = (20 \cdot 15)/25$. By the first Euclid theorem we have $DK \cdot DB = AD^{2}$, from which we obtain that the length of $DK$ is $7^{2}/25$; similarly the length of $DL$ is computed as $DC^{2}/DB = 15^{2}/25 = 9$.

Finally, $AC$ can be seen as the diagonal of a rectangle with sides of length $AK + CL$ and $KL$; by the Pythagorean theorem, its length is
$$\sqrt{(AK + CL)^{2} + (DL - DK)^{2}} = \sqrt{\left(\frac{168}{25} + 12\right)^{2} + \left(9 - \frac{49}{25}\right)^{2}} = 20.$$

As in the previous solutions, one must first show that $ABCD$ can be inscribed in a circle. Having shown this, it is possible to conclude directly thanks to Ptolemy's theorem, which states that a convex quadrilateral $ABCD$ can be inscribed in a circle if and only if $AB \cdot CD + BC \cdot AD = AC \cdot BD$. From this we obtain that the length of $AC$ is $20$.