Math Olympiad Marathon

Problem 1

  1. Reply only answer or solution
  2. All can send me any category questions privately I will post it here

20230227_230332

In the picture AO=BO And \frac {1}{2}AO=\frac{1}{2}BO
RU=6, So what is the area of \triangle{AOB} ?

Last date of replying answer is 03/03/2023 at 9:00 a.m

Ans will given in 5hat day also

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In the question, you have to tell the area of \triangle{AOB}. NOT ''A''.

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āĻ•ā§āĻˇā§‡āĻ¤ā§āĻ°āĻĢāĻ˛ āĻŦā§‡āĻ° āĻ•āĻ°āĻ¤ā§‡ āĻšāĻŦā§‡

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\fbox{You are right.}

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That's right. :+1::+1::+1::+1::ok_hand:

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Problem 2

\fbox{You should give the full process}

1000^n+283+928+2000n+500 āĻ•ā§‡ 1000 āĻĻā§āĻŦāĻžāĻ°āĻž āĻ­āĻžāĻ— āĻ•āĻ°āĻ˛ā§‡ āĻ­āĻžāĻ—āĻļā§‡āĻˇ āĻ•āĻ¤ āĻĨāĻžāĻ•āĻŦā§‡ ?

Last date of posting answer is 06/03/2023 at
12:00 p.m.

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āĻĒā§āĻ°āĻĻāĻ¤ā§āĻ¤ āĻ°āĻžāĻļāĻŋāĻ¤ā§‡ āĻĒā§āĻ°āĻĨāĻŽ āĻ“ āĻšāĻ¤ā§āĻ°ā§āĻĨ āĻ°āĻžāĻļāĻŋ 1000^n āĻ“ 2000n, 1000 āĻĻā§āĻŦāĻžāĻ°āĻž āĻŦāĻŋāĻ­āĻžāĻœā§āĻ¯āĨ¤ āĻ¸ā§āĻ¤āĻ°āĻžāĻ‚ 283+928+500 āĻ•ā§‡ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāĻ˛ā§‡ āĻ¯āĻ¤ āĻ­āĻžāĻ—āĻļā§‡āĻˇ āĻĨāĻžāĻ•ā§‡ āĻ¤āĻžāĻ‡ āĻ‰āĻ¤ā§āĻ¤āĻ°āĨ¤
283+928+500=1711 āĻ•ā§‡ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāĻ˛ā§‡ āĻ­āĻžāĻ—āĻļā§‡āĻˇ 711
āĻ¨āĻŋāĻ°ā§āĻŖā§‡ā§Ÿ āĻ‰āĻ¤ā§āĻ¤āĻ° 711

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By A I meant area, but it seems like I was wrong.

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\fbox{Oh!!!!! I see}

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Your answer is correct.

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Problem 2
Solution

āĻĒā§āĻ°āĻĻāĻ¤ā§āĻ¤ āĻ°āĻžāĻļāĻŋāĻ¤ā§‡ āĻĒā§āĻ°āĻĨāĻŽ āĻ“ āĻšāĻ¤ā§āĻ°ā§āĻĨ āĻ°āĻžāĻļāĻŋ 1000^n āĻ“ 2000n, 1000 āĻĻā§āĻŦāĻžāĻ°āĻž āĻŦāĻŋāĻ­āĻžāĻœā§āĻ¯āĨ¤ āĻ¸ā§āĻ¤āĻ°āĻžāĻ‚ 283+928+500 āĻ•ā§‡ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāĻ˛ā§‡ āĻ¯āĻ¤ āĻ­āĻžāĻ—āĻļā§‡āĻˇ āĻĨāĻžāĻ•ā§‡ āĻ¤āĻžāĻ‡ āĻ‰āĻ¤ā§āĻ¤āĻ°āĨ¤
283+928+500=1711 āĻ•ā§‡ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāĻ˛ā§‡ āĻ­āĻžāĻ—āĻļā§‡āĻˇ 711
āĻ¨āĻŋāĻ°ā§āĻŖā§‡ā§Ÿ āĻ‰āĻ¤ā§āĻ¤āĻ° 711

@Arko_1729 give the 3^{rd} problem

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Problem 3

āĻ¯āĻĻāĻŋ n āĻāĻ•āĻŸāĻž āĻ§āĻ¨āĻžāĻ¤ā§āĻŽāĻ• āĻĒā§‚āĻ°ā§āĻ¨āĻ¸āĻ‚āĻ–ā§āĻ¯āĻž āĻšā§Ÿ, āĻ¤āĻžāĻšāĻ˛ā§‡ n/n+675 āĻ•ā§‡ āĻ•āĻžāĻŸāĻžāĻ•āĻžāĻŸāĻŋ āĻ•āĻ°ā§‡ āĻ˛āĻ˜āĻŋāĻˇā§āĻ  āĻ†āĻ•āĻžāĻ°ā§‡ āĻ˛āĻŋāĻ–āĻ˛ā§‡ p/q āĻšā§ŸāĨ¤
(q-p) āĻāĻ° āĻ¸āĻŽā§āĻ­āĻžāĻŦā§āĻ¯ āĻ¸āĻ•āĻ˛ āĻ­āĻŋāĻ¨ā§āĻ¨ āĻ­āĻŋāĻ¨ā§āĻ¨ āĻŽāĻžāĻ¨ā§‡āĻ° āĻ¯ā§‹āĻ—āĻĢāĻ˛ āĻ•āĻ¤?

āĻāĻ•āĻŸāĻŋ āĻ­āĻ—ā§āĻ¨āĻžāĻ‚āĻļāĻ•ā§‡ āĻ•āĻžāĻŸāĻžāĻ•āĻžāĻŸāĻŋ āĻ•āĻ°āĻ˛ā§‡ āĻ†āĻŽāĻ°āĻž āĻšāĻ° āĻ“ āĻ˛āĻŦāĻ•ā§‡ āĻ¤āĻžāĻĻā§‡āĻ° āĻ—āĻ¸āĻžāĻ—ā§ āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāĻ‡āĨ¤āĻ•āĻžāĻŸāĻžāĻ•āĻžāĻŸāĻŋāĻ° āĻĒāĻ° āĻ˛āĻŦ āĻāĻŦāĻ‚ āĻšāĻ°ā§‡āĻ° āĻ¸āĻžāĻ§āĻžāĻ°āĻŖ āĻ—ā§āĻ¨āĻŋāĻ¤āĻ• 1 āĻšā§ŸāĨ¤

BdMO 2020 national primary

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Problem 4
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .

Last date of posting answer is 13/03/2023 at
12:00 p.m.

Source

IMO 2007

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Problem 4
Solution 1

Let and . Then by the isosceles triangles manifest in the figure we have and , so and . Furthermore and .

If , then . But also , so by SSA “Incongruence” (aka. the Law of Sines:

) we have . This translates into , or , which incidentally equals , as desired.

If , then also

by the Exterior Angle Theorem, so and hence and are supplementary. A simple Law of Sines calculation then gives , as desired. This completes both directions of the proof.

Solution 2

Since , , and similarly, . Since , by considering triangles we have

. It follows that .

Now, by the Law of Sines,

.

It follows that if and only if

.

Since ,

,

and

.

From these inequalities, we see that if and only if (i.e., ) or (i.e., ). But if , then triangles are congruent and , a contradiction. Thus we conclude that if and only if. \alpha+\beta =\cfrac{\pi}{3} Q.E.D.

Problem 5
Let and be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both and ?)

  • A. 2
  • B.4
  • C. 5
  • D.6
  • E. 8

0 voters

Last date of replying answer 18/03/23

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Ok. There are 5 winners.

Problem 5
Solution

There are two radius 3 circles to which and are both externally tangent. One touches the tops of and and extends upward, and the other the other touches the bottoms and extends downward. There are also two radius 3 circles to which and are both internally tangent, one touching the tops and encircling downward, and the other touching the bottoms and encircling upward. There are two radius 3 circles passing through the point where and are tangent. For one is internally tangent and is externally tangent, and for the other is externally tangent and is internally tangent.

Source

AIME QUESTION SOLUTION.

Problem 6
āĻ°āĻžāĻšā§āĻ˛ āĻ¸ā§āĻĨāĻžāĻ¨āĻžāĻ‚āĻ• āĻ¤āĻ˛ā§‡ (3,3) āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ¤ā§‡ āĻ†āĻ›ā§‡āĨ¤ āĻ¸ā§‡ āĻāĻ•āĻ§āĻžāĻĒā§‡ āĻšāĻ¯āĻŧ āĻ¤āĻžāĻ° āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ° āĻāĻ•āĻ˜āĻ° āĻ‰āĻĒāĻ°ā§‡āĻ° āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ¤ā§‡ āĻ¯ā§‡āĻ¤ā§‡ āĻĒāĻžāĻ°ā§‡ āĻ…āĻĨāĻŦāĻž āĻāĻ•āĻ˜āĻ° āĻĄāĻžāĻ¨ā§‡āĻ° āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ¤ā§‡ āĻ¯ā§‡āĻ¤ā§‡ āĻĒāĻžāĻ°ā§‡āĨ¤ āĻ¤āĻžāĻ° āĻŽā§ŒāĻ˛āĻŋāĻ• āĻ¸āĻ‚āĻ–ā§āĻ¯āĻž āĻ–ā§āĻŦāĻ‡ āĻĒāĻ›āĻ¨ā§āĻĻ, āĻ¤āĻžāĻ‡ āĻ¸ā§‡ āĻ•āĻ–āĻ¨ā§‹ āĻāĻŽāĻ¨ āĻ•ā§‹āĻ¨ā§‹ āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ¤ā§‡ āĻ¯āĻžāĻŦā§‡ āĻ¨āĻž āĻ¯āĻžāĻ° āĻ­ā§āĻœ āĻ†āĻ° āĻ•ā§‹āĻŸāĻŋ āĻ‰āĻ­āĻ¯āĻŧāĻ‡ āĻ¯ā§ŒāĻ—āĻŋāĻ•āĨ¤ āĻ¸ā§‡ āĻ•āĻ¤āĻ­āĻžāĻŦā§‡ (20,13) āĻŦāĻŋāĻ¨ā§āĻĻā§āĻ¤ā§‡ āĻĒā§ŒāĻāĻ›āĻžāĻ¤ā§‡ āĻĒāĻžāĻ°ā§‡?


Rahul is at (3,3) on the coordinate plane. In each step, he can move one point up or one point to the right. He loves primes, and will never visit a coordinate point where both values are composite. In how many ways can he reach (20,13) ?