**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 \((\frac{BD}{DC} = \frac{BF}{FA})\) 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 \((\frac{BD}{DC} = \frac{BF}{FA})\) 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 \((\frac{BD}{DC} = \frac{BF}{FA})\)

=> DF||AC

=> ∠FDA = ∠CAD (alternate angles)

Since by (i) ∠FBK = ∠FDK

=> ∠ABE = ∠FBK = ∠FDK = ∠FDA = ∠CAD;

=> ∠ABE = ∠CAD;

**Q.E.D.**

**2. Let \((a_1 , a_2 , … , a_{2011})\) be a permutation of the numbers 1, 2, …, 2011. Show that there exists two numbers j and k such that \((1\le j < k \le 2011)\) and \(|(a_j – j)| = |(a_k -k)|\)**

**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

\((a_1^2 + a_2^2 + … + a_{2011}^2 + 1^2 + 2^2 + … 2011^2 = 2 a_1 1 + 2 a_2 2 + … + 2 a_{2011} 2011 + (0^2 + 1^2 + … + 2010^2))\)

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 \(t_j\) -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 \((m^2) and ((m+1)^2)\).

=> n – k = \((m^2)\) **… (i)**

n + ℓ = \(((m+1)^2)\) **… (ii)**

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

=> 2n + ℓ – k = \((2m^2 + 2m + 1)\)

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

Now(n-k)(n+ℓ) = \((m^2 \times (m+1)^2)\)

Hence \(((m^2 + m)^2)\)= \((n^2)\) + nℓ – nk – ℓk

= \((n^2)\) + n(ℓ-1+1) – nk – ℓk

= \((n^2)\) + n(ℓ-1) – nk + n – ℓk

= n(n + ℓ – 1 – k) + n – ℓk

= n(n + ℓ – k – 1) + n – ℓk

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

Thus

n – ℓk

= \(((m^2 + m)^2 – n(2m^2 + 2m – n))\)

\(((m^2 + m)^2 – 2n(m^2 + m) + n^2)\)

n – ℓk = $latex (m^2 + m – n)^2)$

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

**4. K is a 20-sided convex polygon with vertices \((A_1 , A_2 , … , A_20 )\) 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 \((A_1 A_2) , (A_4 A_5)\) 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. \((BB_1)\) and \((CC_1)\) respectively are the bisectors of ∠B and ∠C with \((B_1)\) on AC and \((C_1)\) on AB. Let E and F be the feet of the perpendiculars drawn from A onto \((BB_1)\) and \((CC_1)\) 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. **\((BB_1)\)** and \((CC_1) \) respectively are the bisectors of ∠B and ∠C with \((B_1) \) on AC and \((C_1) \) on AB. Let E anf F be the feet of the perpendiculars drawn from A onto **\((BB_1)\)** and \((CC_1) \) respectively. Suppose D is the point at which the incircle of ABC touches AB.

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

**Construction:** Let **\((BB_1)\)** and \((CC_1) \) 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. **

**6. Find all pairs \(((x,y) \in \mathbb{R} )\) 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