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

## Thursday, July 30, 2009

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

Labels:
circle,
inscribed,
square,
tangency point

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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

ReplyDeleteVihaan: vihaanup@gmail.com

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).

ReplyDeleteIsn't the distance formula in analytical geometry nothing but simple Pythagoras?

ReplyDeleteVihaan

It is well known that :

ReplyDeleteIf 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