British Mathematics Olympiad (BMO) Round 1 2012

Regional Mathematics Olympiad Region 2 Questions

RMO 2012 solution to Question No. 6

6. Find all positive integers n such that \(3^{2n} + 3 n^2 + 7 \) is a perfect square.

Solution:

We use the fact that between square of two consecutive numbers there exist no perfect square. That is between \(k^2 \) and \((k+1)^2 \) there is no square.

Note that \(3^{2n} = (9^n)^2 \) and  \((9^n + 1)^2 \) are two consecutive perfect square and \(3^{2n} + 3 n^2 + 7 \) is always a number between them for n > 2 (easily proved by induction).

Hence the only solution is n = 2.

RMO 2012 solution to Question No. 5

5. Let ABC be a triangle. Let D, E be points on the segment BC such that BD = DE = EC. Let F be the mid point of AC. Let BF intersect AD in P and AE in Q respectively. Determine the ratio of triangle APQ to that of the quadrilateral PDEQ.

Solution:

Applying Menelaus' theorem to ΔBCF with AD as the transversal, we have
 \(\frac {BD}{DC} \frac{CA}{AF} \frac {FP}{PB} = 1 \)
But BD/DC = 1/2 (as BD = DE = EC) and CA/AF = 2/1 (as CF = FA).
Hence we have BP = PF.
Again applying Menelaus' Theorem to ΔBCF with AE as the transversal we have
\(\frac {BE}{EC} \frac{CA}{AF} \frac {FQ}{QB} = 1 \)
But BE/EC = 2/1 and CA/AF = 2/1
Hence 4FQ = QB.

Suppose FQ= x unit. The QB = 4x unit. That is BF = 5x unit. Since BP = PF hence each is 2.5x unit.
Thus PQ = 2.5x - x = 1.5x unit

Hence  \(\frac{ΔAPQ}{ΔABF} = \frac{1.5x}{5x}\)
Also \(\frac{ΔABF}{ΔABC} =\frac{1}{2}\)

Thus  \(\frac{ΔAPQ}{ΔABF} \frac{ΔABF}{ΔABC}= \frac{1.5x}{5x} \frac{1}{2}\)
=> \(\frac{ΔAPQ}{ΔABC} = \frac{1.5}{10}\) ...(1)

Again \(\frac{ΔADE}{ΔABC} = \frac{1}{3}\) (as DE/BC = 1/3)
Thus \(\frac{ΔADE}{ΔABC} - \frac{ΔAPQ}{ΔABC}  = \frac{1}{3} -  \frac{1.5}{10}\)
=> \(\frac{PQED}{ΔABC}  = \frac{5.5}{30}\) ...(2)

Using (1) and (2) we have 

 \(\frac{ΔAPQ}{PQED} = \frac{4.5}{5.5} = \frac{9}{11}\)

RMO 2012 solution to Question No. 4

4. Let X = {1, 2, 3, ... , 10}. Find the number of pairs {A, B} such that A ⊆ X, B ⊆ X, A ≠ B and A∩B = {5, 7, 8}.

Solution:

First we put 5, 7, 8 in each of A and B.

We are left out with 7 elements of X.

For each of these 7 elements there are three choices:

a) it goes to A

b) it goes to B

c) it goes to neither A nor B

Hence there are total \(3^7\) = 2187 choices. From these 2187 cases we delete that one case where all of the seven elements goes to neither A nor B as A≠ B thus giving 2187 -1 = 2186 cases.

Since A and B is unordered (that is A= {5, 7, 8, 1, 2} , B = {5, 7, 8, 4} is the same as B= {5, 7, 8, 1, 2} , A = {5, 7, 8, 4} ) we take half of these 2186 cases that is 1093 cases.

Hence there are 1093 such pairs.

RMO 2012 solution to Question No. 3

3. Let a and b are positive real numbers such that a+b = 1. Prove that \(a^a  b^b + a^b b^a \le 1\)

Solution:

We use the weighted A.M.-G.M. inequality which states that:

\(\frac {w_1 a_1 + w_2 a_2 }{w_1 + w_2} \ge ({a_1}^{w_1} {a_2}^{w_2})^{\frac{1}{w_1 + w_2}} \)

First we put \(w_1 = a , a_1 = a , w_2 = b, a_2 = b \)
Hence we get \(\frac {a a + b b }{a + b} \ge (a^a b^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(a^2 + b^2 \ge (a^a b^b) \) ....(1)

Similarly we put \(w_1 = a , a_1 = b , w_2 = b, a_2 = a \)

Hence we get \(\frac {a b + b a }{a + b} \ge (b^a a^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(2ab \ge (b^a a^b) \) ....(2)

Adding (1) and (2) we have 

\(a^2 + b^2 + 2ab \ge a^a b^b + b^a a^b \)
=> \((a+b)^2 \ge a^a b^b + b^a a^b \)
As a+b =1 we have the desired inequality
\(1 \ge a^a b^b + b^a a^b \).

RMO 2012 Solution to Question No. 2

2. Let a, b, c be positive integers such that a divides \(b^5\) , b divides \(c^5\) and c divides \(a^5\). Prove that abc divides \((a+b+c)^{31}\).

Solution:

A general term of the expansion of \((a+b+c)^{31}\) is \(\frac {31!}{p!q!r!} a^p b^q c^r\) where p+q+r = 31 (by multinomial theorem; this may reasoned as following: from 31 factors (a+b+c), choose p factors and from those chosen p factors take out 'a'. From remaining 31-p factors choose q factors and from these chosen q factors take out 'b'. From the remaining r factors take out 'c'.)

Now the terms of the expansion can be of three types:

Case 1

p, q and r are all non-zero. These terms are straight away divisible by abc as all of a, b, and c are present in them.

Case 2

Exactly one of p, q, r is zero and rest two are non-zero. Let us examine the subcase where r=0 and p,q are non zero. Other two subcases will be similar.

Then p+q+0 = 31 or p+q=31

Term of the expansion will have :

\(a^p b^q = ab(a^{p-1} b^{q-1})\)

We will show that \(a^{p-1} b^{q-1}\) is divisible by c where p+q=31

Suppose \((p-1)\ge 5 , (q-1)\ge 5\) then c divides \(a^{p-1}\) as it contains \(a^5\)  and by problem c divides \(a^5\)

Again if p-1 < 5 then \((q-1) \ge 25 \) as p-1 + q-1 = 29 (as p+q = 31)

Now a divides \(b^5 \) or \(a^5\) divides \(b^{25}\). As c divides \(a^5\) and \(a^5\) divides \(b^{25}\) hence c divides \(b^{25}\) implying c divides \(b^{q-1}\) as \((q-1)\ge 25 \)

Case 3.

Exactly two of p, q and r are zero. Let us again examine of the three subcases where q=0, r=0 and p nonzero. Other two subcases will be similar.

Then p = 31.

\(a^{31} = a\times a^5 \times a^{25}\). c divides \(a^5\) and b divides \(c^5\) which divides \(a^{25}\).

Hence we have checked all possible terms and have shown than abc divides each of them.

RMO 2012 solution to Question No. 1

1. Let ABCD be a unit square. Draw a quadrant of a circle with A as the center and B, D as the end points of the arc. Similarly draw a quadrant of a circle with B as the center and A, C as the end points of the arc. Inscribe a circle Γ touching the arcs AC and BD both externally and also touching the side CD.

Solution:

Suppose O is the center of Γ. The point of tangency of Γ with CD be P, with arc AC be R and with BD be Q. 

By symmetry DP = 1/2 unit.

Also note that AQO is a straight line. (If we draw a tangent line through Q, AQ will be perpendicular to the tangent line as A is the center of quadrant ABD. Also OQ will be perpendicular to the tangent line as O is center of Γ. OQ, AQ are perpendicular to same line - the tangent line - at the same point (Q). Hence they are the same straight line).

Length of AQO = AQ + QO = 1 + r (AQ = AB = 1 and let the radius of Γ be r)

Drop a perpendicular OM on AD.

Then OM = DP = 1/2

AM = 1-r.

Applying Pythagoras' Theorem on triangle AQM we have

\( (1-r)^2 + (\frac{1}{2})^2 = (1+r)^2 \)

=> r = 1/16.

Answer: Radius of smaller circle is \(\frac{1}{16}\) unit.

Regional Mathematics Olympiad (RMO) 2012