## Thursday, July 30, 2009

### Problem 331. Square, Point on the Inscribed Circle, Tangency Points

Proposed Problem
Click the figure below to see the complete problem 331 about Square, Point on the Inscribed Circle, Tangency Points.

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Complete Problem 331
Level: High School, SAT Prep, College geometry

1. Let O be (0,0) and the side of the square=2a then A:(-a,-a),B:(-a,a),C:(a,a) & D:(a,-a) while E:(-a.0),F:(0,a),G:(a,0) & H:(0,-a) and the radius of the inscribed circle=a. Now using the distance formula it can be shown that PH^2-PE^2=2a(y-x) while PD^2-PB^2=4a(y-x) from which we deduce that: (PD^2-PB^2)/2 = PH^2-PE^2
Vihaan: vihaanup@gmail.com

2. Assume that P is at x distance from line BC and at y distance from DC and side of square be a. Now if you just apply simple pythagoras and just simlply put the values you will get PH^2 - PE^2 = a(y-x) and also PD^2 - PB^2 = 2a(y-x).

3. Isn't the distance formula in analytical geometry nothing but simple Pythagoras?
Vihaan

4. It is well known that :
If Z is the midpoint of a segment XY with XZ = ZY = a, and P is any point not collinear with X,Z,Y then
PX^2 + PY^2 = 2[PZ^2 + a^2] or equivalently,
2PZ^2 = PX^2 + PY^2 - 2a^2
In the present problem,
let 2k be the length of each side of the square.
For convenience, denote the distances of P from A,B,C,D by a,b,c,d respectively
2PH^2 = a^2 + d^2 - 2k^2,
2PE^2 = a^2 + b^2 - 2k^2,
2(PH^2 - PE^2) = d^2 -b^2 = PD^2 - PB^2 etc