Week 2 Solutions

Answers for wk 2:

1) 3 (Online Problem)
2) $\displaystyle{\frac{1}{7}}$ (Art & Craft of Problem Solving, pg 286)

Solution (1)
At first glance, this problem seems to be unsolvable because of the number of variables. However, if we pay more attention to the problem, we see that the range of the variables is only non-negative integers. This implies that all the terms on the L.H.S. of the equation are non-negative integers. Since the L.H.S can only be 1, this implies that 1 of the terms on the L.H.S. is 1 and others 0.

$\Longrightarrow x^2=1, y^2,z^2,2xyz=0 \Longrightarrow (x,y,z)=(1,0,0)$

Since the expression on the L.H.S is cyclic, we'll have the cyclic solution (x,y,z)=(1,0,0),(0,1,0),(0,0,1).
That gives us 3 solutions to the equation, therefore the answer is 3.

Solution (2)
The diagram for problem 2 is shown below:

It would be impossible to measure the area of triangle GHI by finding the base and height and divide the product by 2. In the question, there is an important piece of information: we know the [ABC]=1. Also, the question mentioned something about ratio. This inspires us to actually use ratio of areas of different triangles in the diagram to find [GHI].

One good way of solving a complicated diagram like this would be to simplify the diagram. Sincere there are actually 3 cevians, we'll remove one of them to simplify the diagram. Then, the diagram would be something like this:

There is actually nothing we can do here (except identifying that [ABD]:[ADC]=1:2 since BD:DC=1:2, which is useless since there is nothing else to compare the results with). By observation, we see that if we connect GC (another useful tool is connecting points) it will create another set of triangles that we can compare the areas with. Thus we connect GC and see what happens:

Then, we see that [GBD]:[GDC]=1:2, and also [GCE]:[GAE]=1:2. I think I forgot to cover this in the 'Area' post, but I'll mention this anyway. When we compare areas, we tend to put in variables like x,y to make the areas easier to compare. Thus, [GBD]=x,[GDC]=2x,[GEC]=y,[GAE]=2y:

Time to bring in other sets of areas to compare. Notice that previously we have [ABD]:[ADC]=1:2, and since we know that [ADC]=3y+2x, we get

$\displaystyle{[ABD]=\frac{3y+2x}{2}=\frac{3y}{2}+x \Longrightarrow [ABG]=\left(\frac{3y}{2}+x\right)-x=\frac{3y}{2}}$ :

Now that we found the ratio of [ABG] to the other triangles, it will be more easier to compare. We see that $\displaystyle{[ABG]:[AGE]=\frac{3}{2}:2=3:4 \Longrightarrow BG:GE=3:4}$ .

Since we also have $\displaystyle{[CEB]:[ABE]=1:2 \Longrightarrow 2(3x+y)=\frac{3y}{2}+2y=\frac{7y}{2}}$ , simplifying gives us

$4(3x+y)=7y$
$y=4x$

Converting all y to x gives us the diagram:

It is now easy to see that $[ABG]:[GBD]=6x:1x=6:1 \Longrightarrow GA:GD=6:1$ .

Since we found all the areas in terms of x, it's time to add the one more cevian back in to the diagram. It can be quite confusing so cross-refer the diagram below to the diagram above:

Similarly, we derive the ratio EH:HB=1:6. Since $[ECB]=7x \Longrightarrow [ECH]:[HCB]=x:6x$ .

Since [HCB]=3x+[HGC]=6x,
therefore [HGC]=3x and [EHC]=x.
$\Longrightarrow EH:HG:GB=1:3:3$

Similarly, we see that each cevian is divided into the ratio of 1:3:3.

Notice that G,H,I is the midpoint of the lines HB,CI,AG respectively. This inspires us to join AH,BI:

Since we know the ratios of the lines, it is easy to determine the ratio of areas and after some calculations we will derive the following diagram above.

We need to find [GHI], which is very easy after determining all of the triangles' areas in terms of x.

$\displaystyle{\therefore [GHI]=\frac{3x}{21x}\cdot [ABC]=\frac{1}{7}\cdot [ABC]=\frac{1}{7}}$