SA Mathematics Olympiad (2015) - 3 Tough Questions

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?

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Pretty much a lucky guess. We see that its the only solution to $xyz = 120$ and $(x-2)(y-2)(z-2) = 24$.

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.

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.

Repeatedly using the fact that the midpoint of a triangle divides the triangle into $2$ triangles of equal area should get you the answer.

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 $[DEF] = \frac{1}{8}[ABC] = 3$

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!

The answer is 3.
Since nothing is given for simplicity consider the triangle is equilateral,find out the the square of side(yes side length not required), Now agian for simplicity consider the AB as base of triangle with vertex A at origin and we are done just find out coordinates of D,E,F in terms of 'a' and you are done!

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## Comments

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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,

$2x = 60$

$x=\boxed{30}$

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A really good solution! Good job!

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I think the answer of Q.19 is c) 30°.

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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 $148$.

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

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Pretty much a lucky guess. We see that its the only solution to $xyz = 120$ and $(x-2)(y-2)(z-2) = 24$.

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.

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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.

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Repeatedly using the fact that the midpoint of a triangle divides the triangle into $2$ triangles of equal area should get you the answer.

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 $[DEF] = \frac{1}{8}[ABC] = 3$

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yupp.. nice and easy.. without complications

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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!

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The answer is 3. Since nothing is given for simplicity consider the triangle is equilateral,find out the the square of side(yes side length not required), Now agian for simplicity consider the AB as base of triangle with vertex A at origin and we are done just find out coordinates of D,E,F in terms of 'a' and you are done!

image

Sorry for my poor handwriting :p

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