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

1 Like

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

2 Likes

āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻŦ⧇āϰ āĻ•āϰāϤ⧇ āĻšāĻŦ⧇

1 Like

\fbox{You are right.}

2 Likes

Problem 1
Solution
20230303_093322

In the picture,
RQ=3
\implies\sqrt{9}
\implies\sqrt{5^2-4^2} [a=\sqrt{c^2-b^2}]
So,
OQ=4
OR=5
QR=3

\frac{1}{2}AO=5
Again,\frac{1}{2}AO=\frac{1}{2}OB=5
SO, AO=BO=5+5=10
The area of \triangle{AOB} is =(\frac{1}{2}×10×10)
\implies\frac{100}{2}

\implies\fbox{50} (ANS)

1 Like

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

1 Like

Problem 2

\fbox{You should give the full process}

1000^n+283+928+2000n+500 āϕ⧇ 1000 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ āĻ•āϤ āĻĨāĻžāĻ•āĻŦ⧇ ?

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

āĻĒā§āϰāĻĻāĻ¤ā§āϤ āϰāĻžāĻļāĻŋāϤ⧇ āĻĒā§āϰāĻĨāĻŽ āĻ“ āϚāϤ⧁āĻ°ā§āĻĨ āϰāĻžāĻļāĻŋ 1000^n āĻ“ 2000n, 1000 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝāĨ¤ āϏ⧁āϤāϰāĻžāĻ‚ 283+928+500 āϕ⧇ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāϞ⧇ āϝāϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻĨāĻžāϕ⧇ āϤāĻžāχ āωāĻ¤ā§āϤāϰāĨ¤
283+928+500=1711 āϕ⧇ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāϞ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ 711
āύāĻŋāĻ°ā§āĻŖā§‡ā§Ÿ āωāĻ¤ā§āϤāϰ 711

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

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

1 Like

Your answer is correct.

Problem 2
Solution

āĻĒā§āϰāĻĻāĻ¤ā§āϤ āϰāĻžāĻļāĻŋāϤ⧇ āĻĒā§āϰāĻĨāĻŽ āĻ“ āϚāϤ⧁āĻ°ā§āĻĨ āϰāĻžāĻļāĻŋ 1000^n āĻ“ 2000n, 1000 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝāĨ¤ āϏ⧁āϤāϰāĻžāĻ‚ 283+928+500 āϕ⧇ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāϞ⧇ āϝāϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻĨāĻžāϕ⧇ āϤāĻžāχ āωāĻ¤ā§āϤāϰāĨ¤
283+928+500=1711 āϕ⧇ 1000 āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāϞ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ 711
āύāĻŋāĻ°ā§āĻŖā§‡ā§Ÿ āωāĻ¤ā§āϤāϰ 711

@Arko_1729 give the 3^{rd} problem

Problem 3

āϝāĻĻāĻŋ n āĻāĻ•āϟāĻž āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻĒā§‚āĻ°ā§āύāϏāĻ‚āĻ–ā§āϝāĻž āĻšā§Ÿ, āϤāĻžāĻšāϞ⧇ n/n+675 āϕ⧇ āĻ•āĻžāϟāĻžāĻ•āĻžāϟāĻŋ āĻ•āϰ⧇ āϞāϘāĻŋāĻˇā§āĻ  āφāĻ•āĻžāϰ⧇ āϞāĻŋāĻ–āϞ⧇ p/q āĻšā§ŸāĨ¤
(q-p) āĻāϰ āϏāĻŽā§āĻ­āĻžāĻŦā§āϝ āϏāĻ•āϞ āĻ­āĻŋāĻ¨ā§āύ āĻ­āĻŋāĻ¨ā§āύ āĻŽāĻžāύ⧇āϰ āϝ⧋āĻ—āĻĢāϞ āĻ•āϤ?

āĻāĻ•āϟāĻŋ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāϕ⧇ āĻ•āĻžāϟāĻžāĻ•āĻžāϟāĻŋ āĻ•āϰāϞ⧇ āφāĻŽāϰāĻž āĻšāϰ āĻ“ āϞāĻŦāϕ⧇ āϤāĻžāĻĻ⧇āϰ āĻ—āϏāĻžāϗ⧁ āĻĻāĻŋā§Ÿā§‡ āĻ­āĻžāĻ— āĻĻāĻŋāχāĨ¤āĻ•āĻžāϟāĻžāĻ•āĻžāϟāĻŋāϰ āĻĒāϰ āϞāĻŦ āĻāĻŦāĻ‚ āĻšāϰ⧇āϰ āϏāĻžāϧāĻžāϰāĻŖ āϗ⧁āύāĻŋāϤāĻ• 1 āĻšā§ŸāĨ¤

BdMO 2020 national primary

Problem 3
Solution

At first, P=\frac {n}{gcd(n,n+675)}
and. Q=\frac {675}{gcd(n,n+675)}
So, q-p={\frac {n}{gcd(n,n+675)}}-{\frac{675}{gcd(n,n+675)}}
\implies{q-p=\frac{675}{gcd(n+675)}}
Finally,
All possible value of q-p is
\sum _{d|675}\frac{675}{d}
\implies\sum _{d|675}d
\implies\fbox {1240}

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

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

1 Like

This post was flagged by the community and is temporarily hidden.