Search found 172 matches
- Sun Feb 12, 2012 9:11 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Junior 5
- Replies: 5
- Views: 5742
BdMO National 2012: Junior 5
Problem 5: $ABC$ is a right triangle with hypotenuse $AC$. $D$ is the midpoint of $AC$. $E$ is a point on the extension of $BD$. The perpendicular drawn on $BC$ from $E$ intersects $AC$ at $F$ and $BC$ at $G.$ (a) Prove that, if $DEF$ is an equilateral triangle then $\angle ACB = 30^0$. (b) Prove t...
- Sun Feb 12, 2012 9:09 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Junior 4, Primary 8
- Replies: 8
- Views: 9005
BdMO National 2012: Junior 4, Primary 8
Problem 4: A magic box takes two numbers. If we can obtain the first number by multiplying the second number with itself several times then a green light on the box turns on. Otherwise, a red light turns on. For example, if you enter $16$ and $2$ then the green light turns on because $2×2× 2×2 = 16...
- Sun Feb 12, 2012 9:05 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Junior 2, Primary 3
- Replies: 2
- Views: 3548
BdMO National 2012: Junior 2, Primary 3
Problem:
Prove that, the difference between two prime numbers larger than $2$ can’t be a prime number other than $2$.
Prove that, the difference between two prime numbers larger than $2$ can’t be a prime number other than $2$.
- Sun Feb 12, 2012 8:55 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Secondary 9
- Replies: 6
- Views: 5614
BdMO National 2012: Secondary 9
Problem 9:
Consider a $n×n$ grid of points. Prove that no matter how we choose $2n-1$ points from these, there will always be a right triangle with vertices among these $2n-1$ points.
Consider a $n×n$ grid of points. Prove that no matter how we choose $2n-1$ points from these, there will always be a right triangle with vertices among these $2n-1$ points.
- Sun Feb 12, 2012 8:51 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Secondary 8
- Replies: 10
- Views: 8031
BdMO National 2012: Secondary 8
Problem 8: The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $AB$ and arc $AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point ...
- Sun Feb 12, 2012 8:48 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Secondary 4, Junior 8
- Replies: 3
- Views: 3599
BdMO National 2012: Secondary 4, Junior 8
Problem:
Find the total number of the triangles whose all the sides are integer and longest side is of $100$ in length. If the similar clause is applied for the isosceles triangle then what will be the total number of triangles?
Find the total number of the triangles whose all the sides are integer and longest side is of $100$ in length. If the similar clause is applied for the isosceles triangle then what will be the total number of triangles?
- Sun Feb 12, 2012 8:46 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Secondary 2, Junior 3, Primary 7
- Replies: 7
- Views: 6367
BdMO National 2012: Secondary 2, Junior 3, Primary 7
Problem : When Tanvir climbed the Tajingdong mountain, on his way to the top he saw it was raining $11$ times. At Tajindong, on a rainy day, it rains either in the morning or in the afternoon; but it never rains twice in the same day. On his way, Tanvir spent $16$ mornings and $13$ afternoons witho...
- Tue Jan 31, 2012 11:04 am
- Forum: Algebra
- Topic: Functional equation
- Replies: 10
- Views: 6342
Re: Functional equation
Am I missing something?
What about this one? $f(x)= \frac{1}{2} + \frac{\sqrt{2}}{4},\ \ \forall x$
What about this one? $f(x)= \frac{1}{2} + \frac{\sqrt{2}}{4},\ \ \forall x$
- Tue Jan 31, 2012 10:33 am
- Forum: Algebra
- Topic: Functional equation
- Replies: 10
- Views: 6342
Re: Functional equation
such 'a' function or 'all' functions...?
- Fri May 27, 2011 7:27 am
- Forum: Secondary Level
- Topic: sin s are rational, so..cos s are rational
- Replies: 1
- Views: 2020
sin s are rational, so..cos s are rational
$sin\ A,sin\ B,sin\ C$ of a triangle are rational. Prove that $cos\ A,cos\ B,cos\ C$ are also rational.