## Sunday, May 30, 2010

### Problem 463: Three Concentric Circles, Secant, Tangent, Triangle, Area

Geometry Problem
Click the figure below to see the complete problem 463 about Three Concentric Circles, Secant, Tangent, Triangle, Area. Complete Problem 463

Level: High School, SAT Prep, College geometry

1. Let AD=d. If we consider O to be the origin and x & y axes to the left and down resply. then we can write the following eqns. for the three tangents FA, FB & EC: dx+cy=d^2+c^2, (d-a)x+cy=(d-a)^2+c^2 and (d-a-b)x+cy=(d-a-b)^2+c^2. Solve simultaneously two at a time to obtain:
Area of Tr. FGE=(1/2c)[(2d-a)(-a^2-ab+ad)+(2d-2a-b)(bd)+(2d-a-b)(a^2+ab-ad-bd)] and this simplifies to - ab(a+b)/2c Or the absolute value of Area Tr. FGE = ab(a+b)/2c
Ajit

2. Just want to add here that area of a triangle formed by P1:(x1,y1),P2:(x2,y2) & P3:(x3,y3) is given by (1/2)[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] and may turn out to be +ve or -ve depending on how points 1, 2 & 3 are selected. Hence only the absolute value of the area is really meaningful.
Ajit

3. Typo Correction:
4. 