Geometry (finding angle)?

Question

ABCD is a square. P is a point in the square such that PA = a, PB = 2a and PC = 3a. Find angle APB.

A) 120 deg
B) 130 deg
C) 135 deg
D) 140 deg
E) 145 deg


This question came from a local mathematical olympiad competition I had just a few weeks ago. I have finished it already but I am unsure of the answers. Note that no calculators and geometrical instruments are allowed.

Please show how you do it.

Answer 1

ok, I did it, but I'll warn you that my solution is pretty hardcore. maybe you can find a way to simplify the process.

let d be the side length of the square. there are three interesting triangles that we can form:
1) APB, which has side lengths a, 2a, and d
2) BPC, which has side lengths 2a, 3a, and d
3) APC, which has side lengths a, 3a, and √2*d

for all of these, we can use the law of cosines to get the internal angles α = angle APB and β = angle BPC. note that angle APC = α + β.

the law of cosines, applied to each of the above triangles, gives us the following equations:
1) d^2 = a^2 + (2a)^2 - 2(a)(2a)cos(α)
2) d^2 = (3a)^2 + (2a)^2 - 2(3a)(2a)cos(β)
3) 2d^2 = a^2 + (3a)^2 - 2(a)(3a)cos(α+β)

in each case, we can solve for d^2/a^2 (which we'll henceforth call R):
1) R = 5 - 4cos(α)
2) R = 13 - 12cos(β)
3) R = 5 - 3cos(α+β)

now we'll use the angle addition formula to expand out the last equation in terms of cos(α) and cos(β):
R = 5 - 3[cos(α)cos(β) - √(1-cos^2(α))√(1-cos^2(β))]

on the right hand side, make the substitutions
cos(α) = (5-R)/4
cos(β) = (13-R)/12
to get a big equation where R is the only variable.

after doing a decent amount of algebra you can get the equation to be
√[(R^2 - 10R + 9)(R^2 - 26R + 25)] = -R^2 + 2R + 15.
square both sides and bring all the terms to one side, and what you're left with is
32R^3 - 320R^2 + 544R = 0.
divide by 32R:
R^2 - 10R + 17 = 0.

whew! a quadratic! the solution is
R = 5 ± 2√2,
so
cos(α) = (5-R)/4
= ± √2/2,
which means either α = 45 or α = 135.

Answer 2

the answer is 135, but I'm not sure of it either

draw a vertical line to line AB, the point on AB is signed as E, then draw another vertical line on BC, the point is F.

so PE=PF, since PB=2a,

PB^2 = BE^2 + PE^2
4a^2= BE^2 + PE^2

and AP^2 = PE^2 +AE^2
it's a^2 = PE^2 +AE^2

so 4a^2= BE^2 + PE^2, a^2 = PE^2 +AE^2

BE^2= 3AE^2
BE= √3AE

the next steps are easier, I'm just not sure the answer is right, but the solution should be right.

Answer 3

if you write your finished work ,then i can check it and can say.