INMO Practice Test-I

Time: 4 Hours [100 marks]

The real numbers a, b, c, d satisfy simultaneously the equations

\(abc − d = 1, bcd − a = 2, cda − b = 3, dab − c = −6.\)

Prove that \(a + b + c + d \neq 0.\) [16]Consider a triangle ABC and let M be the midpoint of the side BC.

Suppose \(∠MAC = ∠ABC\) and \(∠BAM = 105◦\) . Find the measure of \(∠ABC.\) [16]There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they both play in the best possible way?[17]

Find all positive integers n such that \(36^n − 6\) is a product of two or more consecutive positive integers. [17]

Consider a triangle \(ABC\) with \(∠ACB = 90◦\) . Let F be the foot of the altitude from C. Circle \(w\) touches the line segment \(FB\) at point \(P\), the altitude \(CF\) at point Q and the circumcircle of ABC at point R. Prove that points \(P, Q, R\) are collinear and \(AP = AC\). [17]

The real positive numbers \(x, y, z\) satisfy the relations \(x ≤ 2, y ≤ 3, x + y + z = 11.\) Prove that \(\sqrt { xyz } ≤ 6.\)

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## Comments

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TopNewestThese questions are not of even RMO level

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BTW, I am Anurag.

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6^2n - 6 = 6( 6^(2n-1) -1).

C1: 6 and 6^(2n-1)-1 are consecutive. In this case, 6^(2n-1)-1 has to be 5 and hence n=1. C2: 6^(2n-1)-1 is the product of chain of consecutive numbers. All the numbers have to be odd, because the product is odd. But if there are two numbers, there is atleast one even number. So, this case is not possible. Therefore, no solution. C3: 6 and 6^(2n-1)-1 is a product of chain of consecutive numbers. We see again, because of case 2, we are limited to the case that 6^(2n-1) is further expanded into chain of consecutive numbers. C4: 3

2(6^(2n-1)-1) is a chain of consecutive numbers. we see again 6^(2n-1)-1 has to be broken down, but its factors when multiplied by 3 or 2 gives some consecutive numbers. We again can see there has to be one even number and again when it is multiplied by 2, the product is divisible by 4, a contradiction since 4 does not divide the product. Hence n=1 is the only solution.Log in to reply

has anyone solved 5th

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Sorry for being too late to post the solution Q.2) Use sine rules for triangles AMC and AMB. Equate the value of AM from both these expressions. Cancel out CM and MB as they both are equal. Now you are left with a simple trigonometric equation in sine(angle ABC). Solve that you will get angle ABC as 30 degrees. Please tell me if the answer is right or wrong

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That's absolutely correct. It is just a little laborious but easy.

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done! no 6 easy!

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A hint for 6th question

Use AM-GM on x/2,y/3 and z/6.

A hint for second question:

Prove that triangles MAC and ABC are similar .

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tnks for the hints

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Thanks for the 6th!

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Can you put answers to them

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Sketch of proof for 3: Prove that all losing positions for a person \(p\) are consecutive/same integers whose sum is divisible by \(3\). Thus, since \(A\) starts with \(2010,2010\) he would lose. We can prove this by providing a suitable algorithm for \(B\) and proving that \((2,1)\) is a losing position.

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PLZ give the diagram of question 5!

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Where did you get this from?

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Woot...Got it...you found it on BPRIM website right? :P

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Yeah, could you post the solutions to some of these questions? I am trying, but this is terribly difficult.

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