Finding the Length of QP in a Circle-Triangle Problem

[utkarshakash]

Homework Statement

Q)A circle C whose radius is 1 unit touches the x-axis at 'A'.The centre Q lies in 1st quadrant.The tangent(other than x-axis) from origin touches the circle at T and a point P lies on it such that OAP is a right angled triangle with right angle at 'A' and its perimeter is 8 units.Then the length of QP is:-
a)0.5
b)4/3
c)5/3
d)None of these

Homework Equations

The Attempt at a Solution

This seems to difficult to me as nothing is given except the radius of the circle. Suppose I assume the point T to be (α,β) I can write the equation of tangent in terms of α and β if only I know the equation of the circle.

 

[verty]

Here is how I would go about it. Let OA = a be given, then find the angle AOQ and AOP. Do some trig to find the 3 sides, you know they add up to 8. Hopefully at that point it'll all work out.

 

[junaid314159]

Here is a general outline of the approach I took:
- Draw a good sketch of the picture
- Draw auxiliary line QT which forms right triangle PTQ (note that right triangle QTP is similar to right triangle OAP). Using similar triangles will make the algebra much easier.
- I let my variables be theta for angle QOA and let segment OA have length 'a', as verty mentioned.
- Then if you work at it (using similar triangles and a little trig) you can show that segment TP has length 2a/(a^2-1). Similarly you can show that segment QP has length (a^2+1)/(a^2-1).
- We now have the lengths of the pieces of the perimiter of right triangle OAP all in terms of a single variable 'a'.
- In other words: 2a + 2a/(a^2-1) + (a^2+1)/(a^2-1) + 1 = 8 for which you can easily solve for a, and then substitute to get the length of QP.

Junaid Mansuri

 

[junaid314159]

There's actually a very elegant way of solving this problem if you can realize that right triangle QTP is not only similar to right triangle OAP but that the base of right triangle OAP has length 'a' and the base of right triangle QTP has corresponding length 1. This means that right triangle OAP is 'a' times bigger than right triangle QTP which means the perimeter of right triangle QTP is simply 8/a.
This leads immediately to the equation: 2a+1+8/a - 1 = 8 which gives you a = 2 without having to do any trig. So the perimeter of triangle QTP must be 4. Then you can use similar triangles to solve for QP.

 

[utkarshakash]

There's actually a very elegant way of solving this problem if you can realize that right triangle QTP is not only similar to right triangle OAP but that the base of right triangle OAP has length 'a' and the base of right triangle QTP has corresponding length 1. This means that right triangle OAP is 'a' times bigger than right triangle QTP which means the perimeter of right triangle QTP is simply 8/a.
This leads immediately to the equation: 2a+1+8/a - 1 = 8 which gives you a = 2 without having to do any trig. So the perimeter of triangle QTP must be 4. Then you can use similar triangles to solve for QP.

Thanks. i used your equation for finding a but the other two equations were different from yours. I used the fact that area of ΔOAP is sum of areas of ΔOAQ, ΔOQT and ΔQTP and the perimeter is equal to 8. That saved a lot of time. I want to know how you proved that QTP and OAP are similar.

 

[junaid314159]

Using areas is a great idea. I didn't try that. Very nice.

They are similar because they are both right triangles and they both contain the same angle, angle P.