# Pre - rmo Mumbai region

Hello people! Today I gave my pre rmo in mumbai. The paper was pretty easy and I am confident of 14-15 right answers. But I have a few doubts and would love if anyone could solve them:

1) In rectangle ABCD, AB=8 and BC=20. Let P be a point on AD such that angle BPC=90. If r1, r2, r3 are the radii of incircles of triangles APB, BPC and CPD, what is the value of r1 + r2 + r3?

2) Let a,b and c be real numbers such that a - 7b + 8c = 4 and 8a + 4b - c = 7. What is the value of a^2 - b^2 + c^2?

3) The circle C1 touches the circle C2 internally at P. The centre O of C2 is outside C1. Let XY be a diameter of C2 which is also tangent to C1. Assume PY>PX. Let PY intersect C1 at Z. If YZ = 2PZ, what is the magnitude of angle PYX in degrees?

Thank You! Note by Aaryaman Gupta
4 years, 8 months ago

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formula of inradius of a right triangle is= 1/2(p+b-h)

in △APB, r1=(AP+AB−PB)/2

Similarly, r2=(PB+PC−BC)/2

And, r3=(PD+CD−PC)/2

∴r1+r2+r3=(AP+AB−PB)+(PB+PC−BC)+(PD+CD−PC)/2

- 2 years, 9 months ago

1. This is an AMC 10 problem. I'll quickly go through the solution here.
Rearrange the equations to become a+8c=4+7b and 8a−c=7−4b

. Now square both equations to get:

a2+16ac+64c2=16+56b+49b2,

and 64a2−16ac+c2=16b2−56b+49.

Adding the two equations, we get 65a2+65c2=65b2+65 . So 65a2−65b2+65c2=65⇒a2−b2+c2=1

.

This is a pretty decent manipulation problem.

- 2 years, 9 months ago

good and easy questions

- 2 years, 9 months ago

but tricky

- 2 years, 9 months ago