Archive for December, 2009

Math: Problem of the day-2

Dec 18th

Posted by admin in Algebra

3 comments

This problem was asked in IMO 1970, Hungary.

Q: For what natural numbers n can the product of some of the numbers n,n+1,n+2,n+3,n+4,n+5 be equal to the product of the remaining ones?

Note that since there are 6 numbers under consideration, if a prime number p divides any of these numbers, it cannot be greater than 5.

In the set of numbers n+1,n+2,n+3,n+4, there can be no common prime divisor greater than 2 or 3. Also, since two of these four numbers must be odd, they must be powers of 3. But this isn’t possible since no two powers of 3 differ by 2.

Therefore, there is no such n

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Math: Problem of the day-1

Dec 17th

Posted by admin in Geometry

4 comments

This problem was asked in IMO 1961, Hungary.

Q: Let a,b,c be the lengths of a triangle ABC whose area is S. Prove that a^2+b^2+c^2 \geq 4S \sqrt{3}. In what case does equality hold?

Solution: We can express the area in terms of the sides and angles of the triangle:

S=\frac{1}{2}bc sin A

Also, the cosine rule says that

a^2=b^2+c^2-2bc cos A

Therefore,

a^2+b^2+c^2 \geq 4S \sqrt{3} \iff b^2+c^2 \geq bc( \sqrt3 sin A + cos A) \\ \iff (b-c)^2+2bc(cos (A-60^0) \geq 0

Note that equality will hold above if b=c and a=60^0, that is, when the triangle is equilateral.

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