olymmath

ZA-SAMO 2002

South African Mathematics Olympiad 2002.

Other problems

Problem 1

Given a quadrilateral ABCDABCD such that AB2+CD2=AD2+BC2AB^2 + CD^2 = AD^2 + BC^2 , prove that AC⊥BDAC \perp BD .

Problem 2

Find all triples of natural numbers (a,b,c)(a,b,c) such that aa , bb and cc are in geometric progression and a+b+c=111a + b + c = 111 .

Problem 3

A small square PQRSPQRS is contained in a big square. Produce PQPQ to AA , QRQR to BB , RSRS to CC and SPSP to DD so that AA , BB , CC and DD lie on the four sides of the large square in order, produced if necessary. Prove that AC=BDAC = BD and AC⊥BDAC \perp BD .

Problem 4

How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem, abcabc is not considered different from bacbac , etc.)

Problem 5

In acute-angled triangle ABCABC , a semicircle with radius rar_a is constructed with its base on BCBC and tangent to the other two sides. rbr_b and rcr_c are defined similarly. rr is the radius of the incircle of ABCABC . Show that 2r=1ra+1rb+1rc.\frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}.

Problem 6

Find all rational numbers aa , bb , cc and dd such that 8a2−3b2+5c2+16d2−10ab+42cd+18a+22b−2c−54d=42,8a^2 - 3b^2 + 5c^2 + 16d^2 - 10ab + 42cd + 18a + 22b - 2c - 54d = 42, 15a2−3b2+21c2−5d2+4ab+32cd−28a+14b−54c−52d=−22.15a^2 - 3b^2 + 21c^2 - 5d^2 + 4ab +32cd - 28a + 14b - 54c - 52d = -22.