RMO 2010

Don't you think that the paper was too easy. I mean no doubt people might have done even all correct.
The cutoffs gonna be high.

I got
Ans3→657

Ans4→{2,3,6} (1,2,4 is surely wrong 11 +14=54 which can't be k2 for any integer k

Ans6→43

Ans1→ simple use of formula area=12ab sinθ

Ans2→ solving eqn. in a,b,c by cramers rule and bit of determinants at the end

Ans5→quad AEMF actually becomes a rhombus so you automatically prove it

3 is definitely 729 and how can 4 be 1,2,4 1/4+1/2=3/4 is not an integral multiple of 1!
first can be solved by taking ratios of areas of triangles in the hexagon as products of their lengths
even i solved 2 by cramers rule there must be a simpler method

on dividing into groups of hundreds you'll get 9 each not 11 wich gives 9into9into9=729
lol if ne1 is interested here are those 729 nos
1116, 1124, 1132, 1148, 1156, 1164, 1172, 1188, 1196, 1212, 1228, 1236, 1244, 1252, 1268, 1276, 1284, 1292, 1316, 1324, 1332, 1348, 1356, 1364, 1372, 1388, 1396, 1412, 1428, 1436, 1444, 1452, 1468, 1476, 1484, 1492, 1516, 1524, 1532, 1548, 1556, 1564, 1572, 1588, 1596, 1612, 1628, 1636, 1644, 1652, 1668, 1676, 1684, 1692, 1716, 1724, 1732, 1748, 1756, 1764, 1772, 1788, 1796, 1812, 1828, 1836, 1844, 1852, 1868, 1876, 1884, 1892, 1916, 1924, 1932, 1948, 1956, 1964, 1972, 1988, 1996, 2116, 2124, 2132, 2148, 2156, 2164, 2172, 2188, 2196, 2212, 2228, 2236, 2244, 2252, 2268, 2276, 2284, 2292, 2316, 2324, 2332, 2348, 2356, 2364, 2372, 2388, 2396, 2412, 2428, 2436, 2444, 2452, 2468, 2476, 2484, 2492, 2516, 2524, 2532, 2548, 2556, 2564, 2572, 2588, 2596, 2612, 2628, 2636, 2644, 2652, 2668, 2676, 2684, 2692, 2716, 2724, 2732, 2748, 2756, 2764, 2772, 2788, 2796, 2812, 2828, 2836, 2844, 2852, 2868, 2876, 2884, 2892, 2916, 2924, 2932, 2948, 2956, 2964, 2972, 2988, 2996, 3116, 3124, 3132, 3148, 3156, 3164, 3172, 3188, 3196, 3212, 3228, 3236, 3244, 3252, 3268, 3276, 3284, 3292, 3316, 3324, 3332, 3348, 3356, 3364, 3372, 3388, 3396, 3412, 3428, 3436, 3444, 3452, 3468, 3476, 3484, 3492, 3516, 3524, 3532, 3548, 3556, 3564, 3572, 3588, 3596, 3612, 3628, 3636, 3644, 3652, 3668, 3676, 3684, 3692, 3716, 3724, 3732, 3748, 3756, 3764, 3772, 3788, 3796, 3812, 3828, 3836, 3844, 3852, 3868, 3876, 3884, 3892, 3916, 3924, 3932, 3948, 3956, 3964, 3972, 3988, 3996, 4116, 4124, 4132, 4148, 4156, 4164, 4172, 4188, 4196, 4212, 4228, 4236, 4244, 4252, 4268, 4276, 4284, 4292, 4316, 4324, 4332, 4348, 4356, 4364, 4372, 4388, 4396, 4412, 4428, 4436, 4444, 4452, 4468, 4476, 4484, 4492, 4516, 4524, 4532, 4548, 4556, 4564, 4572, 4588, 4596, 4612, 4628, 4636, 4644, 4652, 4668, 4676, 4684, 4692, 4716, 4724, 4732, 4748, 4756, 4764, 4772, 4788, 4796, 4812, 4828, 4836, 4844, 4852, 4868, 4876, 4884, 4892, 4916, 4924, 4932, 4948, 4956, 4964, 4972, 4988, 4996, 5116, 5124, 5132, 5148, 5156, 5164, 5172, 5188, 5196, 5212, 5228, 5236, 5244, 5252, 5268, 5276, 5284, 5292, 5316, 5324, 5332, 5348, 5356, 5364, 5372, 5388, 5396, 5412, 5428, 5436, 5444, 5452, 5468, 5476, 5484, 5492, 5516, 5524, 5532, 5548, 5556, 5564, 5572, 5588, 5596, 5612, 5628, 5636, 5644, 5652, 5668, 5676, 5684, 5692, 5716, 5724, 5732, 5748, 5756, 5764, 5772, 5788, 5796, 5812, 5828, 5836, 5844, 5852, 5868, 5876, 5884, 5892, 5916, 5924, 5932, 5948, 5956, 5964, 5972, 5988, 5996, 6116, 6124, 6132, 6148, 6156, 6164, 6172, 6188, 6196, 6212, 6228, 6236, 6244, 6252, 6268, 6276, 6284, 6292, 6316, 6324, 6332, 6348, 6356, 6364, 6372, 6388, 6396, 6412, 6428, 6436, 6444, 6452, 6468, 6476, 6484, 6492, 6516, 6524, 6532, 6548, 6556, 6564, 6572, 6588, 6596, 6612, 6628, 6636, 6644, 6652, 6668, 6676, 6684, 6692, 6716, 6724, 6732, 6748, 6756, 6764, 6772, 6788, 6796, 6812, 6828, 6836, 6844, 6852, 6868, 6876, 6884, 6892, 6916, 6924, 6932, 6948, 6956, 6964, 6972, 6988, 6996, 7116, 7124, 7132, 7148, 7156, 7164, 7172, 7188, 7196, 7212, 7228, 7236, 7244, 7252, 7268, 7276, 7284, 7292, 7316, 7324, 7332, 7348, 7356, 7364, 7372, 7388, 7396, 7412, 7428, 7436, 7444, 7452, 7468, 7476, 7484, 7492, 7516, 7524, 7532, 7548, 7556, 7564, 7572, 7588, 7596, 7612, 7628, 7636, 7644, 7652, 7668, 7676, 7684, 7692, 7716, 7724, 7732, 7748, 7756, 7764, 7772, 7788, 7796, 7812, 7828, 7836, 7844, 7852, 7868, 7876, 7884, 7892, 7916, 7924, 7932, 7948, 7956, 7964, 7972, 7988, 7996, 8116, 8124, 8132, 8148, 8156, 8164, 8172, 8188, 8196, 8212, 8228, 8236, 8244, 8252, 8268, 8276, 8284, 8292, 8316, 8324, 8332, 8348, 8356, 8364, 8372, 8388, 8396, 8412, 8428, 8436, 8444, 8452, 8468, 8476, 8484, 8492, 8516, 8524, 8532, 8548, 8556, 8564, 8572, 8588, 8596, 8612, 8628, 8636, 8644, 8652, 8668, 8676, 8684, 8692, 8716, 8724, 8732, 8748, 8756, 8764, 8772, 8788, 8796, 8812, 8828, 8836, 8844, 8852, 8868, 8876, 8884, 8892, 8916, 8924, 8932, 8948, 8956, 8964, 8972, 8988, 8996, 9116, 9124, 9132, 9148, 9156, 9164, 9172, 9188, 9196, 9212, 9228, 9236, 9244, 9252, 9268, 9276, 9284, 9292, 9316, 9324, 9332, 9348, 9356, 9364, 9372, 9388, 9396, 9412, 9428, 9436, 9444, 9452, 9468, 9476, 9484, 9492, 9516, 9524, 9532, 9548, 9556, 9564, 9572, 9588, 9596, 9612, 9628, 9636, 9644, 9652, 9668, 9676, 9684, 9692, 9716, 9724, 9732, 9748, 9756, 9764, 9772, 9788, 9796, 9812, 9828, 9836, 9844, 9852, 9868, 9876, 9884, 9892, 9916, 9924, 9932, 9948, 9956, 9964, 9972, 9988, 9996,

the more important fact in 5 was to prove that EF bisects AM and is perpendicular to it and it was easy to prove it using properties of triangles

last sum answer is 43 anyways it has to be a silly mistake so check it sol is dead easy

first take
k^2<n<(K+1)^2
therefore
k<√n<k+1

by the prop of greatest integer func root n has to be k
now take n=k^2+1
root n =k
thereofre a=k+1/k
now for n=k^2+2/k therefore a(n+1) is greater or equal to

we can similarly prove for all numbers except for
n=k^2+2k which is geater than k^2+2k+1

so we have now proved this all numbers except k^2-1 are the naswers

so 1775 is the last answer 44^2-1 and 45^2-1 is greater than 2010 so the answer has to
43 as 0 cant be included this is the most efficient proof

4)
sol simple
c=kab/a+b thereofre just use simple hit and trial to get the answer

3 was 729, 4 was 2,3,6, 6 was 43....verify?
much easier than last year this time....silly mistakes in drawing diagrams ruined my proofs

Got the ans as:
3. 729
4. {2,3,6)
6. 43

i should have gone for the rmo everyhting ould have been relaxed for me and i am easily solving 4 of it

Answer of Q4 is 1,2,4 only
Have a look
1 + 1/2 = 2+1/2
= 3/2
= 6 * 1/4
That's an integal multiple of the third number.

And varun.tinkle in th third question it isn't given that alpha is root. It is just given that for x = alpha P1 = P2 = P3

it doesnt matter actually if u just change some concepts and assume it the method will be same

got 4 correct
2) subtracted the equations in cyclic order to get three equation with alpha as root and applied common root and discriminant......
3)729...... simplest of the paper
4)assumed first multiple to be 1 and second to be 2..... and solved .. got the numbers as 2,3,6......
6)43......

will they give ne marks if i wrote the first few steps correctly? anyone getting more than 4 correct?

even I got 4 correct
messed up the 5th and the 3rd one
I hope thats enough for a 12th guy to get through

I don't think that would be enough for a 12th guy.

I tried only 2 and 5

Since i m in 10th so was not able to face the questions 3, 4 and 6

I m not sure bout question 5 also.....

I did 2 in this way..

P1(x) = ax² - bx - c;
P2(x) = bx² - cx - a and,
P3(x) = cx² - ax - b

Now, for α, P1(α) = P2(α) = P3(α)
Case 1. when P1(α) = P2(α)

then, aα² - bα - c = bα² - cα - a
=> α²(a - b) + α (c - b) + (a - c) = 0
=> α²(a - b) + α (c - b) + (a - c) = 0.α² + 0.α + 0
On comparing the coefficients we get,
a - b = 0 , c - b = 0, a - c = 0
=> a = b , b = c, a = c
=> a = b = c
similarly when P2(α) = P3(α) and P3(α) = P4(α) then,
a = b = c

Hence, a = b = c proved....

5. I proved BC = EF