RMO practice sheet - Geometry

 



Note 1: Note the location of P,Q.
P could have been towards AD but if you do that, angle APD would be -40.
Similarly for Q.
Another fact which could have made the solution simpler:
Look at triangle APQ.
There is a point C inside it.
Angle PCQ = 120 = 90 + A/2
So C is the incenter.
So PC,QC are also angle bisectors.













Note that we minimized y + 16 which would also minimize sqrt(y) which is in the numerator.
Isn't that wrong?
No. Since we minimized y + 16 and got it as an expression containing sqrt(y) which got cancelled out and gave us 3/4 which is independent of y.

Another solution sketch.














Solution:
From power of a point:
ZY * ZC = ZG * ZA
ZX * ZB = ZG * ZA
ZB = ZC since AZ is the median.
=> 
ZY * ZB = ZG * ZA
ZX * ZB = ZG * ZA
=> 
ZY * ZB = ZX * ZB
=> ZX = ZY
=> AZ is also the median in triangle AXY.
G already divides AZ in 2:1.
So it's also the centroid for AXY triangle.


Question 16:
Appeared in RMO 2016.

Explanation:
DP bisects ADB because: Triangle: Line perpendicular to internal angle bisector is the external angle bisector
Why is DA = DB?
Since DP is the angle bisector and it meets the circle at M(assume). But M is the same point where the angle bisector of D and perpendicular bisector of AB meet. But perpendicular bisector of AB will pass through P so AK = BK where K is intersection of DP and AB.
Now using angle bisector theorem, DA = DB.





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