Geometry Problem

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

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Complete Problem 463

Level: High School, SAT Prep, College geometry

## Sunday, May 30, 2010

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

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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:

ReplyDeleteF: [(2d-a), (c^-d^2+ad)/c]

G:[(2d-a-b), (c^-d^2+ad)+bd/c]

E:[(2d-2a-b), (-a^2-ab+2ad+c^2-d^2+bd)/c]

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

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.

ReplyDeleteAjit

Typo Correction:

ReplyDeleteF:[(2d-a),(c^2-d^2+ad)/c]

G:[(2d-2a-b),(-a^2-ab+2ad+c^2-d^2+bd)/c]

E:[(2d-a-b),(c^2-d^2+ad+bd)/c]

http://ahmetelmas.files.wordpress.com/2010/06/cemberler-ve-tegetleri.pdf

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