rmo aakash lecture 6

Q1. Let ABCDE be a paper pentagon with AB = AE, angle A = angle B = angle E = 90, BC = 3, CD = 5, and DE = 2. Construct a perpendicular from  A to CD using only a ruler and drawing not more than six lines. All lines have to be drawn inside the pentagon.
S1.
Make A = (0,0) assign co-ordinates and you can get radius r = 6.
Also the circle centered at A with radius AB = r will be touched at B, the tangent from C to B is of length 3.
Similarly the tangent DE = 2.
So the other tangents from D,C to the same circle will have lengths 2,3 respectively.
Using the right triangles ABC, AED, we can show that AC = 3.sqrt(5) and AD = 2.sqrt(10)
Now let K be the point on CD s.t. CK = 3 and KD = 2 (such a point will exist since CD = 5).
In triangle ACD, apply stewart's theorem to get AK = 6.
So, point K will also lie on the circle.
And CD is the tangent to it.

Triangle ABC and AKC are congruent.
So angle ACK = ACB, hence CA is angle bisector of angle BCD.
Similarly AED,AKD are congruent and DA is angle bisector.
Also, since AEB is isosceles, both angles are 45.

Now if we can prove DM and CN are the altitudes in triangle ACD then the third altitude will pass through their intersection which is Orthocenter.

Using co-ordinate geometry compute the intersection points M,N and show that DM and CN are perpendicular to the corresponding sides.

So the 6 lines are:
AC
AD
BE
DM
CN
AH