MHB Find Sum of Diagonals of Pentagon $PQRST$

[anemone]

Let $PQRST$ be a pentagon inscribed in a circle such that $PQ=RS=3$, $QR=ST=10$, and $PT=14$. The sum of the lengths of all diagonals of $PQRST$ equals to $\dfrac{a}{b}$, where $a$ and $b$ are relatively prime positive integers.

Find $a+b$.

 

[Opalg]

View attachment 2752​

[sp]I started with a lucky guess. The diagonals $PR$, $QS$, $RT$ are all equal, because each of them is the longest side of an oblique-angled triangle whose other sides are $3$ and $10$, all three triangles having the same circumradius. The question suggests that the diagonals may all have rational length, and going by the look of the diagram I guessed that those three diagonals might all have length $12$. If so, then the circumradius can be calculated from the formula $$r = \frac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$$ for the circumradius $r$ of a triangle with sides $a,b,c.$ That formula gave the radius of the circle as $\dfrac{72}{\sqrt{95}}.$ I then used the same circumradius formula (this time knowing the radius and two of the sides in order to find the third side) in the triangle $PQS$ to find that $PS = \frac{27}2$, and in the triangle $QST$ to find that $QT = \frac{44}3.$

Since those lengths are all rational, and all consistent with the circumradius being $\frac{72}{\sqrt{95}}$, it follows that my lucky guess was correct. The sum of the lengths of the five diagonals is $12 + 12 + 12 + \frac{27}2 + \frac{44}3 = \frac{385}6$. So the answer to the problem is $385 + 6 = 391.$[/sp]

 

[Albert1]

View attachment 2752​

[sp]I started with a lucky guess. The diagonals $PR$, $QS$, $RT$ are all equal, because each of them is the longest side of an oblique-angled triangle whose other sides are $3$ and $10$, all three triangles having the same circumradius. The question suggests that the diagonals may all have rational length, and going by the look of the diagram I guessed that those three diagonals might all have length $12$. If so, then the circumradius can be calculated from the formula $$r = \frac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$$ for the circumradius $r$ of a triangle with sides $a,b,c.$ That formula gave the radius of the circle as $\dfrac{72}{\sqrt{95}}.$ I then used the same circumradius formula (this time knowing the radius and two of the sides in order to find the third side) in the triangle $PQS$ to find that $PS = \frac{27}2$, and in the triangle $QST$ to find that $QT = \frac{44}3.$

Since those lengths are all rational, and all consistent with the circumradius being $\frac{72}{\sqrt{95}}$, it follows that my lucky guess was correct. The sum of the lengths of the five diagonals is $12 + 12 + 12 + \frac{27}2 + \frac{44}3 = \frac{385}6$. So the answer to the problem is $385 + 6 = 391.$[/sp]

you guessed PR=12
you have to prove:PR=12 first

 

[M R]

By applying Ptolemy's theorem five times I got 5 equations:

$a^2=10c+9$

$a^2=3e+100$

$ce=14a+30$

$ae=3a+140$

$ac=10a+42$

I rerranged a little to get cubics in e-3 and c-10 which my calculator solved for me. :p

 

[anemone]

View attachment 2752​

[sp]I started with a lucky guess. The diagonals $PR$, $QS$, $RT$ are all equal, because each of them is the longest side of an oblique-angled triangle whose other sides are $3$ and $10$, all three triangles having the same circumradius. The question suggests that the diagonals may all have rational length, and going by the look of the diagram I guessed that those three diagonals might all have length $12$. If so, then the circumradius can be calculated from the formula $$r = \frac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$$ for the circumradius $r$ of a triangle with sides $a,b,c.$ That formula gave the radius of the circle as $\dfrac{72}{\sqrt{95}}.$ I then used the same circumradius formula (this time knowing the radius and two of the sides in order to find the third side) in the triangle $PQS$ to find that $PS = \frac{27}2$, and in the triangle $QST$ to find that $QT = \frac{44}3.$

Since those lengths are all rational, and all consistent with the circumradius being $\frac{72}{\sqrt{95}}$, it follows that my lucky guess was correct. The sum of the lengths of the five diagonals is $12 + 12 + 12 + \frac{27}2 + \frac{44}3 = \frac{385}6$. So the answer to the problem is $385 + 6 = 391.$[/sp]

Good job, Opalg!(Happy)

You always be so capable to come up with an educated guess that it couldn't possibly go wrong and what's more important to me is, you will always confirm your hypothesis is correct. :)

By applying Ptolemy's theorem five times I got 5 equations:

$a^2=10c+9$

$a^2=3e+100$

$ce=14a+30$

$ae=3a+140$

$ac=10a+42$

I rerranged a little to get cubics in e-3 and c-10 which my calculator solved for me. :p

Well done, M R! :)

Yes, the trick for solving this problem is to to use the Ptolemy's theorem thrice and thanks for participating!

Here is the solution that I wanted to share with the folks at MHB, which is NOT my solution:

Spoiler

View attachment 2759

Note that $PQRS$ and $QRST$ are isoscles trapezoids so let $d=PR=QS=RT$, $e=PS$ and $f=QT$.

Then by Ptolemy's theorem on $PQRS,\,QRST,\,RSTP$, we have

$d^2=10e+9$

$d^2=3f+10$

$de=10d+42$

Now, solve the first for $e$ and substitute into the third to get:

$d\left(\dfrac{d^2-9}{10}\right)=10d+42$

This is equivalent to the cubic $d^3-109d-420=0$ which has roots 12, -5 and -7 so $d=12$.

Substituting this into the first and second equations gives $e=\dfrac{27}{2}$ and $f=\dfrac{44}{3}$ and thus the sum of all diagonals is $3d+e+f=\dfrac{385}{6}=\dfrac{a}{b}$ and hence $a+b=385+6=391$.

 

[Opalg]

you guessed PR=12
you have to prove:PR=12 first

That is correct. Fortunately, the circumradius formula comes to the rescue again, to justify my "lucky guess".

View attachment 2761​

[sp]Let $x = PR=RT = QS$. The triangles $PQR$ (with sides $3,10,x$) and $PRT$ (with sides $x,x,14$) have the same circumradius. So the circumradius formula says that $$\frac{30x}{\sqrt{(13^2-x^2)(x^2-7^2)}} = \frac{14x^2}{\sqrt{14^2(14^2 - x^2)}}.$$ After some cancellation and simplification, that reduces to $60 = x\sqrt{13^2-x^2}$. Square both sides and rearrange, to get $x^4 - 13x^2 + 60^2 = 0$. That factorises as $(x^2-12^2)(x^2-5^2) = 0$, with solutions $x = \pm5,\pm12.$ The only one of those that fits the geometric context is $x=12.$[/sp]