# Thread: Cyclic quad

1. ## Cyclic quad

Had "fun" with this:

Circle radius 65 has center M(65,65), thus tangent to both x and y positive axis.

ABCD is a cyclic quad, it's vertices on the circumference of circle M.

It's area is 6072 square units.

Coordinates of A are (2,81), coordinates of C are (98,9).

What are the coordinates of B and D?

In case some of you suspect I'm "fishing" for an answer:
the x and y coordinates of B and D add up to 260 (that's from my solution).

I'm posting this for your pleasure(!), hoping nobody finds more than 1 solution

2. ## Re: Cyclic quad

Are the coordinates positive integers or any real numbers is acceptable?

3. ## Re: Cyclic quad

Positive integers.

4. ## Re: Cyclic quad

Answer is:
A(2,81)
B(2,49)
C(98,9)
D(128,81)
But as you can switch B and D there are two answers to this question.

5. ## Re: Cyclic quad

Thanks; mine was:
A(2,81), B(98,121), C(98,9), D(32,9) ..... clockwise !

6. ## Re: Cyclic quad

OK I did a quick calculations on your answer.
So with your points:
Distance between A and B: $\displaystyle a=104$
Distance between B and C: $\displaystyle b=112$
Distance between C and D: $\displaystyle c=66$
Distance between D and E: $\displaystyle d=78$
Using Brahmagupta's formula:
$\displaystyle A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$
Where semiperimeter $\displaystyle s$ is defined as $\displaystyle s=\frac{a+b+c+d}{2}$
So with your numbers, semiperimeter s would be:
$\displaystyle s=\frac{104+112+66+78}{2}=180$
And Area would be:
$\displaystyle A=\sqrt{(180-104)(180-112)(180-66)(180-78)}=7752$
Am I doing something wrong or did you miss something somewhere?

7. ## Re: Cyclic quad

Oh oh...7752 is also what I have for that one; my bad, copied wrong one;
1000 apologies of which you may have 1

8. ## Re: Cyclic quad

No problem.
At least it had one answer... and I had fun solving it anyways.