112) I don't think so.
My proof's a bit weird....shall post if I get a better one.
(1) Let ABC be a triangle with AB =3, BC =4 and CA =5. A Line L,which is perpendicular to AC,Intersects AC in Q and AB in P.Suppose there is a Circle inside the Quadrilateral PBCQ touching all its four sides (i.e, PBCQ has an incircle). Find the area of PBCQ.
(2) Let a and b be 2 natural numbers. Can both a2+4b and b2+4a be perfect squares.
(3) If x,y,z are positive real numbers, prove that x^{2}/y^{2}+y^{2}/z^{2}+z^{2}/x^{2}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}
(4)Consider the following array:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
show that no matter which 5 numbers are removed from this array, there exists a 5-term Arithmetic progression among the remaining 20 numbers
(5) Let ABC be a triangle with circum-circle T. Let M be the point of intersection of this tangent to T at B and C. Suppose N is a point choosen on the line AC such that MN is parallel to AB . prove that NA =NB.
(6) Find all Integers which are measures, in degrees, of the interior angles of the regular polygons.
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7 Answers
11Number 4 does not deserve a place in the RMO paper!
Since 5 nos. are removed in all, 1 has to be from each row.
So now we have to deal with the nos. that are struck out from the 1st row viz..(1,2,3,4,5).
Till 4, we can show that for each number x belonging to row number (say i) – there exists a row number (i.e.={i+1}), satisfying the given conditions.
When 5 is struck out, then it’s same procedure unless and until 9 or 10 is removed from 2nd row.
If it’s 9 or 10, that are removed, then we need to apply the same argument – but this time to the diagonal nos. starting from 1, i.e (1,7,13….) and (2,8,14…..)….from which we can safely prove the existence of such an ap.
115)
Angle BCM=k maan lete hai.
Then angle CBM=k=BAC=MNC.
So CNBM is concyclic.
So angle BNM=angle ABN=k, implying NB=NA.
1(3) Let x/y = a, y/z = b, z/x = c
to prove that a2 + b2 + c2 ≥ a + b + c with abc = 1
AM - GM :
a2 + a2 + a2 + a2 + b2 + c26 ≥ (a2 a2 a2 a2 b2 c2)1/6
4a2 + b2 + c2 ≥ 6a
similarly, a2 + 4b2 + c2 ≥ 6b
a2 + b2 + 4c2 ≥ 6c
add these three inequalities
111(2) clearly a2 + 4b > a2
let us take : a2 + 4b = (a+1)2
if a is odd then it'll become odd+even=even which is not possible
if a is even it'll even+even=odd which is also not possible
so, a2 + 4b ≥ (a+2)2
=> b ≥ a + 1 ......... (1)
similarly, b2 + 4a ≥ (b+2)2
=> a ≥ b + 1 .......... (2)
both (1) and (2) can't hold simultaneously
so no possible integral values of a and b
1What shud be d expected cutoff for class 12 th student in dis paper?
1Regional Mathematical Olympiad-2009
Chhattisgarh Region
Problems and Solutions
1. Let ABC be a triangle with AB = 3, BC = 4 and CA = 5. A line L, which
is perpendicular to AC, intersects AC in Q and AB in P. Suppose there is a
circle inside the quadrilateral PBCQ touching all its four sides(i.e., PBCQ has
an incircle). Find the area of PBCQ.
Solution:
Observe that APQ is similar to ACB. Hence
AP/AC
=AQ/AB
=PQ/BC
=k ,
say. We get AP = 5k, AQ = 3k, PQ = 4k. Thus CQ = 5 − 3k, PB =3− 5k.
Because PBCQ has an incircle, the sides of PBCQ satify the relation
PB + CQ = BC + PQ.
This leads to 4k + 4 = 8 − 8k, giving k= 1/3. Thus
AQ = 3k = 1, PQ = 4k =4/3
.
We get the area of APQ:
[APQ] =
1/2AQ × PQ =2/3
.
Note that the area of ABC is 6. Hence the area of PBCQ is given by:
[PBCQ] = [ABC] − [APQ] = 6 −2/3=16/3
.
2. Let a and b be two natural numbers. Can both a2 + 4b and b2 + 4a be perfect
squares?
Solution: NO! Suppose b <= a. We show that a^2+4b cannot be a square. Observe
that
a^2 < a^2 + 4b <= a^2 + 4a < (a + 2)^2.
This shows that a^2+4b = (a+1^)2. Hence 4b = 2a+1, which is clearly impossible.
Similarly, a<= b implies that b^2 + 4a cannot be a square.
3.
Simplest question in dis paper just appply A.M>=G.M
4. Consider the following 5 × 5 array:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
Show that no matter which 5 numbers are removed from this array, there exists a
5-term arithmetic progression among the remaining 20 numbers.
Solution: Let a, b, c, d, e be the five numbers that are deleted. If some row or
column remains completely in tact in this process, we may take the five numbers
in that particular row or column, which are in AP. Thus we may assume that
a, b, c, d, e belong to different rows as well as different columns. We may further
assume that a, b, c, d, e lie in rows 1,2,3,4,5 respectively. If one of b − a, c − b, d −
c, e − d is at least 6, say b − a>= 6, then there are 5 consecutive numbers between
a and b, which make an AP of 5 numbers. Assume b − a <= 5, c − b <= 5, d − c <= 5
and e − d 5. Since a, b, c, d, e belong to different columns, no difference is equal
to 5. Thus each difference is <= 4. Since a, b, c, d, e are in different rows, this can
happen only if a = 5, b = 9, c = 13, d = 17 and e = 21. Now we can choose
4, 8, 12, 16, 20, which are in AP.
6. Find all integers which are measures, in degrees, of the interior angles of regular
polygons.
Solution: Suppose the regular polygon has n sides. Then the measure of its
interior angle is 180(n − 2)/n, in degrees. If this has to be an integer, n must
divide 360. The divisors of 360 are(there are 24 of them)
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
We have to leave out 1 and 2. For the remaining, we get the angles(22 in all):
60, 90, 108, 120, 135, 140, 144, 150, 156, 160, 162, 165,
168, 170, 171, 172, 174, 175, 176, 177, 178, 179.
