# Problem Of The Week #404 Feb 12th, 2020

Status
Not open for further replies.

#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

-----

Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than 15 such that $AB\cdot DA=BC \cdot CD$. What is the largest possible value of $BD$?

-----

#### anemone

##### MHB POTW Director
Staff member
Congratulations to Opalg for his correct solution, which you can find below:

From the cosine rule in the triangle $BCD$, $$BD^2 = BC^2 + CD^2 - 2BC\cdot CD\cos(\angle BCD).$$ Similarly, in the triangle $BAD$ \begin{aligned}BD^2 &= BA^2 + AD^2 - 2BA\cdot AD\cos(\angle BAD) \\ &= BA^2 + AD^2 + 2BC\cdot CD\cos(\angle BCD) \end{aligned} (because $BA\cdot AD = BC\cdot CD$, and $\cos(\angle BCD) = -\cos(\angle BAD)$ because opposite angles of a cyclic quadrilateral add up to $180^\circ$).

Add those two equations to get $$2BD^2 = BA^2 + AD^2 + BC^2 + CD^2.$$ So to maximise $BD^2$, we want to maximise the sum of the squares of the four sides of the quadrilateral. Thus we want to make those sides as long as possible.

Since the sides are integers less than $15$, the longest possible length for a side is $14$. Suppose that $CD = 14$. Then because $BA\cdot AD = BC\cdot CD$ it follows that $BA$ or $AD$ must be a multiple of $7$. But the only multiple of $7$ that is less than $15$ and distinct from $14$ is $7$ itself. So one of those sides, say $BA$, must be $7$. But then $\dfrac{AD}{BC} = \dfrac{CD}{BA} = \dfrac{14}7 = 2$. So the largest possible values for $AD$ and $BC$ are $12$ and $6$. Then $2BD^2 = BA^2 + AD^2 + BC^2 + CD^2 = 7^2 + 12^2 + 6^2 + 14^2 = 49+144+36+196 = 425$ and so $BD = \sqrt{425/2}\approx 14.577$.

That was on the assumption that one of the sides has length $14$. So we still have to check whether there is a better solution in which the longest side is shorter than that. There cannot be a side of length $13$ because (as in the argument above) one of the other sides would have to be a multiple of $13$, and there are no such multiples less than $15$. In the same way, there cannot be a side of length $11$. So if there is no side of length $14$ then the sum of the squares of the four sides cannot be greater than $12^2 + 10^2 + 9^2 + 8^2 = 144+100+81+64 = 389$, which is less than $425$.

In conclusion, the largest possible value of $BD$ is $\sqrt{425/2}$.

Status
Not open for further replies.