Some challenging maths questions?

Question

1. Two Boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least two of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box have the same colour and have the same size (radius).

2. For all positive real numbers a, b, c prove that
a/(b + c) + b/(a + c) + c/(b + a) ≥ 3/2

3.Find the remainder when 2^1990 is divided by 1990

4. prove that the radius of the excircle opposite angle A of a triangle is (area)/(semiperimeter - BC)

Please try to answer most number of questions and show steps.

gôhpihán · Accepted Answer

1.
I have no idea.

2.
I don't have a good answer, but here's my approach

Proving the inequality is similar to prove that the minimum value of LHS is 3/2

Let k = a + b + c, thus k is strictly positive and greater than a, b, c

So
a/(b + c) + b/(a + c) + c/(a + b)
= a/(k - a) + b/(k - b) + c/(k - c)
Since the expression above is the sum of 3 positive values, so to minimize the expression, we need to maximize the denominator. For this to be done, the denominators must be equal
==> k - a = k - b = k - c
== > a = b = c
==> a/(b + c) + b/(a + c) + c/(a + b) = a/(2a) + b/(2b) + c/(2c) = 1/2 + 1/2 + 1/2 = 3/2

3.
http://answers.yahoo.com/question/index;_ylt=AmbLJuuIlAhJxru_rN1HcUDty6IX;_ylv=3?qid=20110627032530AAc2u8l

4.
http://www.flickr.com/photos/53475957@N06/6097341448/in/photostream

To calculate r_a, let the points of contact of the ecircles with the sides of triangles ABC be P, Q, and R. Denote O and center of ecircle opposite to angle A.

Let PB = x, and PC = y. Then RB = x, and QC = y.

Then from figure in the link, x + y = a, and as the tangents from an external point, in this case A, are equal
c + x = b + y

Solving the equations
x + y = a
x - y = b - c
Simultaneously gives
x = (a + b - c)/2 = s - c, and y = (a - b + c)/2 = s - b

Notice also that AR = c + x = c + (s - c) = s and that AQ = s

Looking at area, and noticing that the area of triangle ABC is
Area of AROQ - Area BROP - area CQOP
= 2 * area ARO - 2 * area BRO - 2 * area CQO
= 2 * (1/2) (r_a * s) - 2 * (1/2) r_a (s - c) - 2 * (1/2) r_a (s - c)
= r_a (s - (s - c) - (s - b))
= r_a * (b + c - s)
= r_a * (s - a)

Thus
Area of triangle = r_a * (s - a)

And r_a = Area of triangle / (s - a)

In this case a is the length BC

(extracted from a trigonometry book)

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Some challenging maths questions?

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