# Find Area of Cyclic Quadrilateral?

Find the minimum area of a cyclic quadrilateral which has integer sides a, b, c and d, integer polygon diagonals p and q, rational circumradius R , and integer area A

Update:

Bonus:

Can you find a solution for minimum area in which all sides have different values, In fact I could find only one such answer :)

Update 2:

Yes Rozeta Area = 234 That's what I too got. BTW I thought about this question when I was solving Dragan's last question.

Update 3:

BTW I have no idea about those thumb downs :)

Update 4:

I have to wait for sometime to see if anyone can beat Rozeta's answer!

Relevance

Vikram, since in a triangle a-b-p we have 4AR = abp the condition that R is rational is implied if all the other elements are rational. Thus, I will assume, as Scythian, that you meant to write integer circumradius R.

Scythian has put two 30-40-50 right triangles together to arrive at an area of 1200.

However, if I take one 30-40-50 right triangle and construct upon the 40-leg a triangle where the other two sides are 14 and 30, then R obviously stays at 25 (since the 30-40-50 is a right triangle), while one diagonal p is 40 (the common side - a leg of the right triangle), and the area is 600 + 168 = 768. The remaining diagonal is either 48 or 40 (this latter case meaning a trapezoid).

But wouldn't it be even more clever to take a 6-8-6-8 rectangle? That has two diagonals of length 10, an R of 5, and an area of 48.

By the way, the first solution that I gave is the smallest non rectangle, non "kite" solution.

The point of Dragan's answer is that none of the angles are right angles.

• I'm going to go with A.

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Edit: Heheh... a thumb's down for the technically correct, but entirely useless answer ;-).

By the way, my answer was [slightly] more than just a smartass answer. I was pointing out that the question needs to be revised to be unambiguous. At the moment, due to the scope of the symbol A, one reading is that we're fixing the area to be some integer A. However, now every cyclic quadrilateral that satisfies the requirements will have area of A.

Something like this works (I believe):

Find the least positive integer A such that there is a cyclic quadrilateral of area A with integral side-lengths, integral diagonals and rational circumradius.

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The only thumb down I see it my own. Earlier it seemed to be glitching and showing others.

• Two right triangles:

1) a=15, b=20, c=25

2) a=7, b=24, c=25

A quadrilateral is formed by these triangles, such that the order of its sides is: 15, 7, 24 and 20. One diagonal is, obviously, 25 and the other is 20, R=12.5 and Area=234. The quadrilateral is cyclic, of course.

http://i299.photobucket.com/albums/mm286/rozeta53/...

Edit:

Who's that 'party breaker' giving TDs, lol?

• I don't know if this is the minimum area, but a cycle quadrilateral of the following dimenions exists:

a = 25

b = 25

c = 25

d = 11

p = q = 30

A = 432

But R is not an integer. Ugh.

Edit: Okay, try this:

a = 30

b = 30

c = 40

d = 40

p = 50

q = 48

R = 25

A = 1200

Well, I'll keep at this one.

Edit: Quadrillerator, I see that you've improved on my answer considerably. In fact, it could be pretty hard to beat.

Edit 2: Okay, hold on for that bonus question.

• For bonus I found:

a=25

b=39

c=60

d=52

p=63 (connect sides a and d)

q=56

A = 1764

R=32.5

If R also need to be integer, than all sides take twice, and area 4 times more.

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Rozeta, very nice! :)