RMO 2011 SOLUTIONS

Solution:

Diagram

Given: ABC be any triangle. AD, BE and CF are drawn from A, B, C to BC, CA and AB respectively such that they concur at K and $latex (\frac{BD}{DC} = \frac{BF}{FA})$ and ∠ADB = ∠AFC.

R.T.P.:  ∠ABE = ∠CAD;

Construction: FD is joined.

Proof: 

Since ∠ADB = ∠AFC; 
hence 180° - ∠BFC = ∠ADB;
=> ∠BFC + ∠ADB = 180°;
=> BDKF is a cyclic quadrilateral (since sum of an opposite pair of angles 180°)

Hence  ∠FBK = ∠FDK (angles in the segment FK in the cyclic quadrilateral BDKF) .... (i)

Since $latex (\frac{BD}{DC} = \frac{BF}{FA})$
=> DF||AC
=> ∠FDA = ∠CAD (alternate angles)
Since  by (i) ∠FBK = ∠FDK
=> ∠ABE = ∠FBK = ∠FDK = ∠FDA = ∠CAD;
=>  ∠ABE = ∠CAD;

Q.E.D.

Solution:  
Let $latex |(a_j - j )| = (t_j)$

If all the $latex t_j$ -s are different then they will take up each value from 0 to 2010 (included) exactly once.


squaring both sides and adding all terms we get

$latex (a_1^2 + a_2^2 + ... + a_{2011}^2 + 1^2 + 2^2 + ... 2011^2  = 2 a_1  1 + 2  a_2 2 + ... + 2  a_{2011} 2011 + (0^2 + 1^2 + ... + 2010^2))$

The left hand side of the equality is even and the right side is odd. Hence all the numbers from 0 to 2010 cannot appear as values of the 2011 $latex t_j$ -s. Hence there must be repetitions.

Proved.

Solution:

The the two consecutive squares be $latex (m^2) and ((m+1)^2)$. 
=> n - k = $latex (m^2)$  ... (i)
n + ℓ = $latex ((m+1)^2)$ ... (ii)
Adding (i) and (ii) we have  (n+ ℓ) + (n-k) = (m^2 + m^2 + 2m + 1)
=> 2n + ℓ - k = $latex (2m^2 + 2m + 1)$
=> ℓ - k = $latex (2m^2 + 2m + 1 - 2n)$  --- (iii)
Now(n-k)(n+ℓ) = $latex (m^2 \times (m+1)^2)$
Hence $latex ((m^2 + m)^2)$= $latex (n^2)$ + nℓ - nk - ℓk
= $latex (n^2)$ + n(ℓ-1+1) - nk - ℓk
= $latex (n^2)$ + n(ℓ-1) - nk + n - ℓk
= n(n + ℓ - 1 - k) + n - ℓk
= n(n + ℓ - k - 1) + n - ℓk
=$latex n(n + 2(m^2) + 2m + 1 - 2n - 1)$ + n - ℓk  .. (replacing ℓ - k by $latex (2m^2 + 2m + 1 - 2n)$ using --- (iii) )
Thus
n - ℓk
= $latex ((m^2 + m)^2 - n(2m^2 + 2m - n))$
$latex ((m^2 + m)^2 - 2n(m^2 + m) + n^2)$

n - ℓk = $latex (m^2 + m - n)^2)$

HENCE PROVED that  n - ℓk is a perfect square.

Solution:

Diagram:


Given: ABC be a triangle. $latex (BB_1)$ and $latex (CC_1) $ respectively are the bisectors of ∠B and ∠C with $latex (B_1) $ on AC and $latex (C_1) $ on AB. Let E anf F be the feet of the perpendiculars drawn from A onto $latex (BB_1)$ and $latex (CC_1) $ respectively. Suppose D is the point at which the incircle of ABC touches AB.

Required to Proof: AD = EF;

Construction: Let $latex (BB_1)$ and $latex (CC_1) $ intersect at I which is then the incentre of the triangle. ID is the perpendicular dropped on AB from I. Then ID is an inradius. EF and AI are joined. Let O be the midpoint of AI. OD, OA, OE and OF are joined.


Proof: 


∠ADI = ∠AEI (by given hypothesis both the right angles)
=> A, I, E, D concyclic (since angles subtended by the segment AI at D and E are equal)
=> Also AI is the diameter
Similarly A, I, E and F are concyclic.
But one and only one circle can pass through three non collinear points. Hence only one circle passes through A, I and E.
Hence all the five points A, I, E, D and F are on the same circle which has AI as the diameter.
Since O is the midpoint of AI and AI is the diameter; hence O is the center of the circle through A, D, E, I, F.

In triangle ΔOAD and ΔOEF
OD = OE (radii of the same circle)
OA = OF (radii of the same circle)
∠AOD = ∠EOF = 180° - A (proof of this claim follows)

Hence ΔOAD  ≅ ΔOEF => AD = EF (HENCE PROVED)


Lemma
R.T.P.: ∠AOD = ∠EOF = 180° - ∠A
Proof: ∠AOD = 2∠AID (angle at the circumference is half the angle at the center).

Also ∠AID = 90° - ∠A/2 (since I is the incenter, AI bisects ∠A; hence in ΔADI, ∠ADI = 90°, ∠DAI = ∠A/2; remaining ∠AID = 90° - ∠A/2)

Hence  ∠AOD = 2∠AID = 2(90° - ∠A/2) = 180° - ∠A -- (*)

Again in  ΔIBC, ∠IBC = ∠B/2; ∠ICB = ∠C/2; hence remaining ∠BIC = 180° - (∠B/2+ ∠C/2) = 90° + ∠A/2; Hence ∠EIF = 90° + ∠A/2 (vertically opposite angles);
Hence ∠EAF = 90°- ∠A/2 (Since EAFI is a cyclic quadrilateral, sum of opposite angles ∠EIF and ∠EAF is 180°)
Thus  ∠EOF = 2∠EAF = 180° - ∠A (Since angle at the center is twice the angle at the circumference) -- (**)

Hence from  (*) and (**) we conclude ∠AOD = ∠EOF = 180° - ∠A

Q.E.D. 

Solution: