The **first** round of the South African Mathematics Olympiad was held today at various high schools throughout the country. Can you solve the two hardest questions in the paper?

**first** round of the South African Mathematics Olympiad was held today at various high schools throughout the country. Can you solve the two hardest questions in the paper?

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TopNewestSolution to QUESTION 19 : ( Its not very complicated)

Let FECG be a kite , then

\(FE=FG and EC=GC\)

\(GC=GA \) (Since diagonals of a rhombus bisect each other)

\(EC=BE\) (Since \(E\) is mid-point of \(BS\)

Therefore, \(BC=AC\)

But, \(AB=BC\) (Sides of a rhombus)

So, \(AB=BC=AC\)

\(\bigtriangleup ABC\) is an equilateral triangle

Since medians of an equilateral triangle also bisect the angle from which they are drawn,

\(2x = 60\)

\(x=\boxed{30}\) – Rishabh Tripathi · 1 year, 7 months ago

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– Mark Mottian · 1 year, 7 months ago

A really good solution! Good job!Log in to reply

This challenging question also featured in the paper:

Triangle ABC has a point D on line AB such that D is the midpoint of AB. E is the midpoint of CD. F is the midpoint of AE. If the area of triangle ABC is 24, find the area of triangle DEF.

– Mark Mottian · 1 year, 7 months agoLog in to reply

Hijacking @@Krishna Sharma's diagram, we have \( [ADC] = \frac{1}{2}[ABC] \), \([ADE] = \frac{1}{2}[ADC] \) and \( [DEF] = \frac{1}{2}[ADE] \). Combining all the equations, we have \( [DEF] = \frac{1}{8}[ABC] = 3 \) – Siddhartha Srivastava · 1 year, 7 months ago

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– Rishabh Tripathi · 1 year, 7 months ago

yupp.. nice and easy.. without complicationsLog in to reply

– Mark Mottian · 1 year, 7 months ago

Hi Siddhartha! This is a very insightful solution. If you knew that the "median of the triangle divides the triangle into two triangles of equal area", it makes the problem very easy. Your solution is definitely elegant. Thank you so much for sharing!Log in to reply

image

Sorry for my poor handwriting :p – Krishna Sharma · 1 year, 7 months ago

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I think the answer of Q.19 is c) 30°. – Dipesh Shivrame · 1 year, 7 months ago

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For 20, I don't see any easier solution other than seeing that a rectangular prism with side lengths \( 6,5,4 \) works and thus the surface area is \( 148 \). – Siddhartha Srivastava · 1 year, 7 months ago

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– Mark Mottian · 1 year, 7 months ago

Siddhartha, how did you figure out that the side lengths were 4,5 and 6?Log in to reply

The first equation is obvious. The second one comes is from the fact that the total volume of the cuboid not colored is \( (x-2)(y-2)(z-2) \), since we have to discount the cubes which appear on either ends. – Siddhartha Srivastava · 1 year, 7 months ago

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