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

ক্ষেত্রফল বের করতে হবে

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 কে 1000 দ্বারা ভাগ করলে ভাগশেষ কত থাকবে ?

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

প্রদত্ত রাশিতে প্রথম ও চতুর্থ রাশি 1000^n ও 2000n, 1000 দ্বারা বিভাজ্য। সুতরাং 283+928+500 কে 1000 দিয়ে ভাগ দিলে যত ভাগশেষ থাকে তাই উত্তর।
283+928+500=1711 কে 1000 দিয়ে ভাগ দিলে ভাগশেষ 711
নির্ণেয় উত্তর 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

প্রদত্ত রাশিতে প্রথম ও চতুর্থ রাশি 1000^n2000n, 1000 দ্বারা বিভাজ্য। সুতরাং 283+928+500 কে 1000 দিয়ে ভাগ দিলে যত ভাগশেষ থাকে তাই উত্তর।
283+928+500=1711 কে 1000 দিয়ে ভাগ দিলে ভাগশেষ 711
নির্ণেয় উত্তর 711

@Arko_1729 give the 3^{rd} problem

Problem 3

যদি n একটা ধনাত্মক পূর্নসংখ্যা হয়, তাহলে n/n+675 কে কাটাকাটি করে লঘিষ্ঠ আকারে লিখলে p/q হয়।
(q-p) এর সম্ভাব্য সকল ভিন্ন ভিন্ন মানের যোগফল কত?

একটি ভগ্নাংশকে কাটাকাটি করলে আমরা হর ও লবকে তাদের গসাগু দিয়ে ভাগ দিই।কাটাকাটির পর লব এবং হরের সাধারণ গুনিতক 1 হয়।

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.