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Circle And Intersection (Posted on 2008-01-25) Difficulty: 3 of 5
P, Q and R are three points located on a circle L with diameter 4 and satisfying PQ = QR. Point S is located inside L in such a manner that QR = RS = SQ. The line passing through P and S intersects L at the point T.

Determine the length of ST.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Attempt | Comment 3 of 6 |

In trying to solve this problem, I found it easier to fix the length of the sides of QRS to each be 2 units.  The problem can be reformulated as follows:

Six points (p, Q, R, S, T, C) have the following properties:
- PQ = QR = RS = QS
- P, Q, R are not colinear
- P and S are on the same side of QR
- P, S, and T are colinear
- CP = CQ = CR = CT

Prove or disprove ST = CT

------

To solve this problem, first fix points Q, R, and S on a coordinate system with Q=(-1,0), R=(1,0), S=(0,sqrt[3])
Let P have coordinates (a,b).
Let T have coordinates (u,v).
Since C is equidistant from Q and R, its coordinates can be expressed as (0,c).
[Note C and T are also on the same side of QR as P and S]

Using distance and slope formulas, the following equations can be formed:
1: (a+1)^2 + b^2 = 4
2: c^2+1 = a^2 + (c-b)^2
3: c^2+1 = u^2 + (c-v)^2
4: (b-sqrt[3])/a = (v-sqrt[3])/u

Similarily the conjecture ST = CT can be expressed as:
H1: u^2 + (c-v)^2 = u^2 + (v-sqrt[3])^2

Equations 1 and 2 can be rewritten as:
1: a^2 + b^2 = 3 - 2a
2: a^2 + b^2 = 1 + 2bc

Equating eqns 1 and 2 yields:
5: 3 - 2a = 1 + 2bc -> a = 1-bc

Substituting eqn 5 into eqn 1 or 2 yields:
6: (2-bc)^2 + b^2 = 4 -> (b)*(b*c^2 +b - 4c) = 0

The factor (b) in eqn 6 is extraneous (it would make P, Q, R colinear) and can be removed, yielding:
7: b = (4c)/(c^2+1)

Substituting eqn 7 back into eqn 5 yields:
8: a = (1-3c^2)/(c^2+1)

Eqn 3 can be solved for u:
9: u = sqrt[1 + 2cv - v^2]

Substituting eqns 7, 8, and 9 into eqn 4:
10: ( (4c)/(c^2+1) - sqrt[3] )/( (1-3c^2)/(c^2+1) ) = ( v - sqrt[3] )/( sqrt[1 + 2cv - v^2] )

Simplifying eqn 10 gives:
11: sqrt[1 + 2cv - v^2] * (4c - (c^2+1)*sqrt[3]) = (v - sqrt[3]) * (1 - 3c^2)

Simplifying eqn H1 gives:
H2: v = (c+sqrt[3])/2

In order for the conjecture ST=CT to be true, the substitution of eqn H2 into eqn 11 must result in 0=0:
sqrt[1 + 2c((c+sqrt[3])/2) - ((c+sqrt[3])/2)^2] * (4c - (c^2+1)*sqrt[3]) = ((c+sqrt[3])/2 - sqrt[3]) * (1 - 3c^2)

sqrt[1/4 + c*sqrt[3]/2 + 3(c^2)/4] * (4c - (c^2+1)*sqrt[3]) = ((c-sqrt[3])/2) * (1 - 3c^2)

(1/2) * sqrt[(1+c*sqrt[3])^2] * (4c - (c^2+1)*sqrt[3]) = (1/2) * (c-sqrt[3]) * (1 - 3c^2)

(1+c*sqrt[3]) * (4c - (c^2+1)*sqrt[3]) = (c-sqrt[3]) * (1 - 3c^2)

-3c^3 + 4*sqrt[3]*c^2 - 3c - sqrt[3]*c^2 + 4c - sqrt[3] = -3c^3 + c + 3*sqrt[3]*c^2 - sqrt[3]

0 = 0

Since the substitution resulted in an identity, the conjecture that ST=CT is true.  In the original problem statement, CT would be a radius, so ST=CT=2.


  Posted by Brian Smith on 2008-01-26 18:11:24
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