**1. Let ABC be a triangle. Let D, E, F be points on the segments BC, CA and AB such that AD, BE and CA concur at K. Suppose and ∠ADB = ∠AFC. Prove that ∠ABE = ∠CAD.**

*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 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

=> DF||AC

=> ∠FDA = ∠CAD (alternate angles)

Since by (i) ∠FBK = ∠FDK

=> ∠ABE = ∠FBK = ∠FDK = ∠FDA = ∠CAD;

=> ∠ABE = ∠CAD;

**Q.E.D.**

**2. Let be a permutation of the numbers 1, 2, …, 2011. Show that there exists two numbers j and k such that and**

**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

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 -s. Hence there must be repetitions.

**Proved. **

**3. A natural number n is chosen strictly between two consecutive perfect squares. The smaller of these two squares is obtained by subtracting k from n and the larger one is obtained by adding ℓ to n. Prove that n – kℓ is a perfect square.**

*Solution:*

The the two consecutive squares be .

=> n – k = **… (i)**

n + ℓ = **… (ii)**

Adding (i) and (ii) we have (n+ ℓ) + (n-k) = (m^2 + m^2 + 2m + 1)

=> 2n + ℓ – k =

=> ℓ – k = $latex (2m^2 + 2m + 1 – 2n)$ **— (iii)**

Now(n-k)(n+ℓ) =

Hence = + nℓ – nk – ℓk

= + n(ℓ-1+1) – nk – ℓk

= + n(ℓ-1) – nk + n – ℓk

= n(n + ℓ – 1 – k) + n – ℓk

= n(n + ℓ – k – 1) + n – ℓk

= + n – ℓk .. (replacing ℓ – k by $latex (2m^2 + 2m + 1 – 2n)$ using **— (iii)** )

Thus

n – ℓk

=

n – ℓk = $latex (m^2 + m – n)^2)$

**HENCE PROVED that n – ℓk is a perfect square.**

**4. K is a 20-sided convex polygon with vertices in that order. Find the number of ways in which three sides of K can be chosen so that every pair among them has at least two sides of K between them. For example and $latex (A_11 A_12)$ is a permissible triplet, but $latex (A_1 A_2) , (A_4 A_5)$ and $latex (A_19 A_20)$ is not.**

*Solution:*

**1. Exactly two sides between one pair of selected sides**

**20 triplets.**

**200 triplets**.

**2. Exactly three sides between one pair of selected sides**

**20 triplets**.

^{rd}side has to be selected from the remaining 7 sides (20- 5 -4 -4) leading to =

**140 triplets**.

**3. Exactly four sides between one pair of selected sides**

**20 triplets**.

^{rd}side has to be selected from the remaining 4 sides (20- 6 -5 -5) leading to = 8

**0 triplets**.

**4. Exactly five sides between one pair of selected sides**

**20 triplets**.

^{rd}side has to be selected from the remaining 1 sides (20- 7 -6 -6) leading to =

**20 triplets**.

**ANSWER: 520**

**5. ABC is a triangle. and respectively are the bisectors of ∠B and ∠C with on AC and on AB. Let E and F be the feet of the perpendiculars drawn from A onto and respectively. Suppose D is the point at which the incircle of ABC touches AB. Prove that AD = EF.**

*Solution:*

**Diagram: **

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

**Required to Proof:** AD = EF;

**Construction:** Let and 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. **

**6. Find all pairs such that $latex (16^{x^2 + y} + 16^{x + y^2})$ =1**

Solution:

hey cheenta …. awesome post ….. BTW what do u expect the cutoff to be ? im expecting around 30

@shivnadimpalli … the cutoff differs from state-to-state… where are you from?

the second problem is simple too: there are 2011 nos, and the difference between any two can range from 1 to 2010. hence, by pigeonhole principle, atleast the differences between two pairs of numbers should be the same. hence proved.:) i got 4 ques right. i hope i get selected!

@ vasudhaThe pigeon hole principle that you have applied does not work here. Think closely and you will see the glitch … but thanks for responding … keep in touch