practice problems

Q1. Let O be the Circumcentre of triangle ABC. Find the value of  angle BAC + angle OBC in degrees.
S1.
Angle OBC = angle OCB since OB = OC in triangle OBC.
2OBC = OBC + OCB = 180 - 2A
=> OBC = 90 - A
angle BAC = A
OBC + BAC = 90 deg = answer

Q2.

Triangle ( AMC ) is isosceles with ( AM = AC ). Medians ( MV ) and ( CU ) are perpendicular to each other, and ( MV = CU = 12 ). What is the area of ( \triangle AMC )?

S2.
Let the centroid be G.
It will divide the triangle into 3 smaller ones with same areas => GCA,GMC,GMA
Area of MGC is easy to find since GM = 2/3*12 = 8 = CG and it is a right triangle.
[MGC] = 1/2 * 8 * 8 = 32
[AMC] = 3*32 = 96 = Answer

Q3.

In a ( \triangle ABC ), let ( O, G, H ) denote its circumcenter, centroid, and orthocenter respectively. If GH/OH = m/n, where ( m, n ) are relatively prime positive integers, find ( m + n ).

S3.
Assume that ABC is not equilateral else it will become 0/0.
Since this question doesn't mention the type of triangle we can use a quick trick to assume right triangle.
In a right triangle:

|\

| \
|   \  O

| G  \
|       \
-------
H
It's clear that OGH is a straight line and HG = 2OG => OH = 3OG => GH/OH = 2/3 => m+n = 5
In fact this is a clever trick to remember Euler line's ratios. Proof here.

Q4.

Solution:
Trick solution:
Assume a scalene right triangle so that angles B and C are unequal. Let A = 90 B = 60 C = 30.
B - C = 30
IAC = 45
O is midpoint of BC and OA = OC => Triangle OAC is isosceles => angle OAC = 30 => IAO = 15.
=> Answer = 1.

Now formal solution.
WLOG Angle B > Angle C so that B - C > 0.
Angle IAC = A/2
Now angle AOC = 2B since angle by chord at center = 2.angle by chord at circumference.
Triangle OAC is isosceles since OA = OC.

=> 2B + 2OAC = 180 => OAC = 90 - B
Angle IAO = IAC - OAC = A/2 - 90 + B = B + A/2 - (A+B+C)/2 = (B-C)/2
So (B-C)/2.IAO = 1 = Answer.