Geometry test - pending Q8 (need to solve it well)

Q1. Let ABC be a given equilateral triangle. Denote the mid-points of sides BC, CA, AB respectively by A1, B1, C1. Three distinct parallel lines p, q,r are drawn through A1, B1, C1, respectively. Line p cuts B1C1 at A2; line q cuts C1A1 at B2; line r cuts A1B1 at C2. Prove that the lines AA2, BB2, CC2 are concurrent.

S1.

Q2.

Let (P) be any point inside triangle ABC. Let A1,A2 in AP, B1,B2 in BP, C1,C2 in CP with

BA1 || AB2 || CP,
CB1 || BC2 || AP,
AC1 || CA2 || BP

Prove that
[A1B1C1] = [A2B2C2]

where ([K]) denotes the area of (K).

S2.

 Q3.

In a triangle ABC, let (AD), (BE), (CF) be the medians. Prove that

AB+BC+CA <= 4/3(AD+BE+CF).
When does equality hold?
S3.

2z + 2y > AB
2z + 2x > AC
2x + 2y > BC
Add all:
4(x+y+z) > (AB + BC + AC)____[1]
LHS = 4/3(AD + BE + CF)
Equality will hold for a degenerate triangle when A,B,C are colinear.
For e.g. let C lie on F, then CF = 0 and centroid also lies on F.
2z = AC, 2y = BC and x = 0

So [1] becomes:
2AC + 2BC + 0 >= 2AB
And both sides are equal. H.P.

Q4.

In triangle ABC, the internal angle bisectors AA1 and BB1 are drawn, with A1, B1 on sides BC, AC

respectively. Prove that the distance from any point M of A1B1 to line AB is equal to the sum

of distances from M to AC and BC.
S4.

First a note on Linear interpolation which we will use here.
Let 3 points A,B,C be colinear s.t. B divides AC internally in ratio t:(1-t).
Let L be a line which is not parallel to ABC s.t. A,B,C are on the same side of L.
Then:
d(L,B) = d(L,A).(1-t) + d(L,C).(t)
where d(L,B) = perpendicular distance of point B from line L.
Proof:

We will use this property in our solution.

Let M divide A1B1 in ratio t:1-t.
We need to prove that:
d(AC,M) + d(BC,M) = d(AB,M)

d(AC,M) = d(AC,A1)(1-t) + d(AC,B1).t ____[1]
d(BC,M) = d(BC,A1)(1-t) + d(BC,B1).t ______[2]
d(AB,M) = d(AB,A1)(1-t) + d(AB,B1).t _______[3]

In [1], d(AC,B1) = 0 => d(AC,M) = d(AC,A1)(1-t)
In[2], d(BC,A1) = 0 => d(BC,M) = d(BC,B1).t

Add [1] and [2]:
d(AC,M) + d(BC,M) = d(AC,A1)(1-t) + d(BC,B1).t
Since A1 lies on angle bisector of A => d(AC,A1) = d(AB,A1)
Similarly d(BC,B1) = d(BA,B1)

Use these and H.P.

Q5.

Let k(a natural number) be a given constant. Initially, there are (n) checkers on the table, where (n > 0). Two persons take turns to remove at least 1 and at most (k) checkers each time from the table. The last person who can remove any checker wins the game.

For what values of (n) will the first person have a winning strategy? For what values of (n) will the second person have a winning strategy? (Give the condition on (n) in terms of (k).)

S6.
My tentative start:

n<=k p1 will win by taking all checkers in one go.

n=k+1, p2 will win as p1 will have to leave <= k checkers and p2 will take them all.

n=k+2 to 2k+1, p1 wins by leaving k+1 checkers for p2.

n = 2k + 2 => p2 wins since p1 can't make it k+1 so p2 will make it k+1 and win.

So multiples of (k+1) => p2 wins and p1 wins otherwise?

Let's try to do it more formally.
Let's say P1 gets n = q(k+1) + r where 0<r<k+1 and q>=1.
P1 removes 'r' and hands over multiple of (k+1) to P2.
P2 can't hand over a multiple of (k+1) to P1 so he will give back q1(k+1) + r1.
If q1 = 0 P1 will take all and win since 1<=r1<=k.
If q1 = 1 then P1 will take r1 and hand over k+1 to P2 who will lose.
and so on.
Since P1 can always hand over a multiple of (k+1) to P2, he will always win.

But if n = q(k+1) right a the beginning then P1 gets trapped and P2 will win from there using the same logic.


Q6.

S6.
This is from UKMOG 2016 p4.

Now add all X,Y components to get the co-ordinates of X, assuming (0,0) is Y.
Co-ordinates of X will be:
-4 -sqrt(3) -sqrt(2).cos(15) ,  2 + sqrt(3) + sqrt(2).sin(15)
Use cos(15) = [sqrt(6) + sqrt(2)]/4 
Use sin(15) = [sqrt(6) - sqrt(2)]/4 

And XY will eventually become: 3 + 3.sqrt(3)

Another solution:

Q8.

From vertex C of the right angled (triangle ABC) (right-angled at (C)), height (CK) is dropped, and in triangle ACK internal angle bisector (CE) is drawn. The line passing through point (B) parallel to (CE) meets (CK) at point (F). Prove that line (EF) divides segment (AC) into equal halves.

S8.

Triangle FKB ~ CKE => FK/CK = KB/KE = FB/CE
EBC is isosceles => CB = EB

Q9.

Two players play a game on an m × n chocolate bar made up of small squares. The players

take turns choosing a square and eating it, together with all the squares below it and to the

right. The top left square is poisoned: the player who eats it, loses. Show that the first player

has a winning strategy. (Assume m, n > 1)

S9.
We just need to show that the strategy exists for player 1. We don't need to show what that strategy is.
We will prove by contradiction.
Let's say player 2 has a winning strategy.
So for example player 1 plays a simple move, i.e. to take the bottom right square.
Now let's say player 2 has a winning strategy, and he picks square (i,j) and all the squares below and towards the right.
But if picking (i,j) leads to a win, player 1 would have made that move and it would have taken the bottom right square along with it.
So contradiction.

It's called a Non-constructive proof where we don't construct the winning strategy but show that it exists for a player.
This game is known as Chomp.

Q10. Square PQRS is inscribed into triangle ABC so that vertices P and Q lie on sides AB and AC and

vertices R and S lie on BC. Express the length of the square’s side in terms of a and ha, where

a, ha denote the lengths of side BC and altitude from A to BC respectively.

S10.

Triangle ABD ~ PBS
=>
PB/AB = PS/AD = k/ha where k is the square side.
Also:
APQ ~ ABC
=> AP/AB = PQ/BC = k/a
=> k/a + k/ha = 1
=> k = a.ha/(a + ha) = answer.