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 $latex (3^{2n} + 3 n^2 + 7 )$ 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
$latex (\frac {BD}{DC} \frac {CA}{AF} \frac {FP}{PB})$ = 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 $latex (\frac {BE}{EC} \frac {CA}{AF} \frac {FQ}{QB})$ = 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 $latex (\frac {\triangle APQ}{\triangle ABF})$ = $latex (\frac {1.5x}{5x})$
Also $latex (\frac {\triangle ABF}{\triangle ABC} = \frac {1}{2})$
Thus $latex (\frac {\triangle APQ}{\triangle ABF}$ = $latex (\frac {\triangle ABF}{\triangle ABC})$ = $ \frac {1.5x}{5x} \frac {1}{2}$ = $latex (\frac {\triangle APQ}{\triangle ABC})$ = $latex (\frac {1.5}{10})$ ...(1)

Again $ \frac {\triangle ADE}{\triangle ABC}$ = $latex (\frac {1}{3})$ (as DE/BC = 1/3)
Thus $ \frac {\triangle ADE}{\triangle ABC} - \frac {\triangle APQ}{\triangle ABC} = \frac {1}{3} - \frac {1.5}{10}$
$latex (\frac {PQED}{\triangle ABC} = \frac {5.5}{30})$ ...(2)
Using (1) and (2) we have $latex (\frac {\triangle APQ}{PQED} = \frac {4.5}{5.5} = \frac {9}{11})$

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 {aa + bb}{a + b}}\ge (a^a b^b)^{\frac {1}{a + b}}$
As a+b =1
we have $ a^2 + b^2 \ge (a^ab^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

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

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

Complex Numbers versus Projective Geometry - One problem, Two solutions

The Problem

Suppose ABC is any triangle. D, E, F are points on BC, CA, AB respectively such that $latex (\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB})$. Prove that the centroids of triangles ABC and DEF coincide.

A little Complex Number

Let A, B, C be points on the Complex plane with complex coordinates a, b, c.

The origin of the plane is made to coincide with the centroid of triangle ABC (the complex coordinates of the vertices adjusted accordingly).

Hence the complex coordinate of the centroid G is 0.

$latex (G = \frac{a+b+c}{3} = 0)$ implying a+b+c= 0.

Next we find the complex coordinates of D, E, F in terms of a, b, c and the given ratio $latex (\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = k)$ (say)

Using the section formula we find that

$latex (d = \frac{ck+b}{k+1})$
$latex (e = \frac{ak+c}{k+1})$
$latex (f = \frac{bk+a}{k+1})$

where d, e, f are the complex coordinates of D, E, F respectively.

Hence the centroid of triangle DEF is given by $latex (\frac{d+e+f}{3} = \frac{\frac{ck+b}{k+1} + \frac{ak+c}{k+1} + \frac{bk+a}{k+1} } {3} = 0)$ using (a+b+c) = 0.

Thus the centroid of triangle ABC and DEF are same.

A little Projective Geometry

We project triangle ABC into a plane T such that the projection is an equilateral triangle.

(to be continued)