# Thread: Area of the triangle through the circumcenter

1. ## Area of the triangle through the circumcenter

Tryin' to help my niece on her homework: The area of triangle ABC is 133. The angle bisectors meet at point I. AB:BC:CA=4:7:8. What is the area of triangle BCI?

We have that angle bisectors meet at the circumcenter, which is the center of the circumscribed circle. We have a bunch of things about medians, perpendicular bisectors, and altitudes. It seems logical that BCI is about 1/3 of ABC, but you know the math teacher isn't going to accept that for an answer.

Tx

2. ## Re: Area of the triangle through the circumcenter

Tryin' to help my niece on her homework: The area of triangle ABC is 133. The angle bisectors meet at point I. AB:BC:CA=4:7:8. What is the area of triangle BCI?

We have that angle bisectors meet at the circumcenter, which is the center of the circumscribed circle. <-- unfortunately no
We have a bunch of things about medians, perpendicular bisectors, and altitudes. It seems logical that BCI is about 1/3 of ABC, but you know the math teacher isn't going to accept that for an answer.

Tx
1. The point I is the centre of the inscribed circle (therefore the name I). Let r denote the radius of the incircle.

2. Then the area of the complete triangle is calculated by:

$A_{\Delta} = \frac12 a r+\frac12 b r+\frac12 c r = \frac12 r (a+b+c)$

3. You don't know the actual lengthes of the sides but

c = 4k
a = 7k
b = 8k

That means the area of the triangle is

$A_{\Delta} = \frac12 r k (4+7+8)$

4. The area of the triangle BCI is

$A_{BCI} = \frac12 r k \cdot 7$

5: The proportion $\frac{A_{BCI}}{A_{\Delta}} = \frac{\frac12 r k \cdot 7}{\frac12 r k (4+7+8)} = \frac7{19}$

6. Now you can determine the area of $\Delta(BCI)$. I've got 49.

3. ## Re: Area of the triangle through the circumcenter

Thanks, earboth. Good solution.

Did they really think a 10th grader could do this problem? Wow.