A primer for mathematics competitions - Section 1.2 - Appetizer Problem

Problem:

Solution:

Here I present my solution. The book's solution is MUCH simpler but that is not mine.

We know $BX : XC = 2:3$, so let $\vec{BX} = 2\vec{p}$. $\vec{XC} = 3\vec{p}$.
Similarly, $\vec{CY} = \vec{q}$, $\vec{YA} = 2\vec{q}$.

Let $\vec{AB} = \vec{r}$, our goal is to compute the length ratios.

$\vec{AX} = \vec{AB} + \vec{BX} = \vec{r} + \vec{2p}$.
$\vec{AX} = \vec{AC} + \vec{CX} = (-3\vec{q}) - \vec{3p}$.

Therefore: $\vec{AX} = \frac{1}{5}(3\vec{r} - 6\vec{q})$ (Just eliminate $\vec{p}$ from the equations above)

$\vec{BY} = \vec{BA} + \vec{AY} = -\vec{r} - 2\vec{q}$

Observe that we have two ways to get to $O$, either starting from $A$ or $B$.

Let $\vec{AO} = \alpha \vec{AX}$
Let $\vec{BO} = \beta \vec{BY}$

Now $\alpha ( \frac{1}{5}(3\vec{r} - 6\vec{q})) = \alpha \vec{AX} = \vec{AO} = \vec{AB} + \vec{BO} = \vec{r} + \beta \vec{BY} = \vec{r} + \beta(-\vec{r} - 2\vec{q})$

Now the amazing thing happens, we can solve $\alpha$ and $\beta$ at the same time by equating coefficients.

$\alpha ( \frac{1}{5}(3\vec{r} - 6\vec{q})) = \vec{r} + \beta(-\vec{r} - 2\vec{q})$

$\frac{3}{5}\alpha = 1 - \beta$ (matching coefficients of $\vec{r}$)
$-\frac{6}{5}\alpha = -2\beta$ (matching coefficients of $\vec{q}$)

Solving, get $\alpha = \frac{5}{6}$, $\beta = \frac{1}{2}$.

So $AO : OX = 5 : 1$, $BO : OY = 1 : 1$.

Next, we tackle the $CO:OZ$ and $AZ : ZB$. To do that, we compute $CO$ in two ways.

Let $\vec{CO} = \gamma \vec{CZ}$
Let $\vec{AZ} = \delta \vec{AB}$

$\vec{CO} = \vec{CB} + \vec{BO} = \vec{CA} + \vec{AB} +  \frac{1}{2}\vec{BY} = 3\vec{q} + \vec{r} + \frac{1}{2}(-\vec{r} - 2\vec{q}) = \frac{1}{2}(\vec{r} + 4\vec{q})$

$\vec{CZ} = \vec{CA} + \vec{AZ} = \vec{3q} + \delta \vec{AB} = \vec{3q} + \delta\vec{r}$.

Now $\frac{1}{2}(\vec{r} + 4\vec{q}) \vec{CO} = \gamma \vec{CZ} = \gamma(\vec{3q} + \delta\vec{r})$

Now we equate coefficients again and get

$\frac{1}{2} = \gamma \delta$ (matching coefficients of $\vec{r}$)
$2 = 3\gamma$

Solving, get $\gamma = \frac{2}{3}$, $\delta = \frac{3}{4}$.

So $CG : GZ = 2 : 1$, $AZ:ZB = 1 : 3$.

Now at this point, we have all the length ratios we wanted. For triangles with opposite angle, we know triangle area is $\frac{1}{2}ab\sin\theta$, so the length ratio give us triangle area ratios.

Since we do not need to vectors anymore, abusing the notation, we reuse $P Q R S T U$ to mean the triangle areas of $\triangle AOZ, \triangle AOY, \triangle ZOB, \triangle YOC, \triangle BOX, \triangle XOC$.

The length ratios give us these equations:

$2P = 5U$
$R = 2S$
$5T = Q$

We wanted $R + T$, but we do not know $Q$, so even more is required. Here is another observation, $\triangle ABX : \triangle ACX = 2:3$ because they share the same height but with different base, that gives $3(P + R + T) = 2(Q + S + U)$, at this point, we can solve $Q = \frac{5.5U + 4S}{1.4}$, and therefore $R + T = 2a + \frac{16.5b + 4a}{7}$.

Even then, it is not enough, because we have no matching answer, now we pull in one more equation with the observation that $\triangle CAE : \triangle CBE = 3:1$ using the similar argument, now we get 5 equations:

$2P = 5U$
$R = 2S$
$5T = Q$
$3(P + R + T) = 2(Q + S + U)$
$P + Q + S = 3(R + T + U)$,

We know these equations will not give an exact answer because it is simply up to scaling. We arbitrarily set $U = 1$ and solve these equations (with a solver, coz I am lazy), we finally get our answers.

P = 2.500
Q = 4.722
R = 0.556
S = 0.278
T = 0.944
U = 1.000

So R + T = 1.5 and that is $\frac{9}{2}\frac{1}{3} = \frac{9b}{2}$

So finally, that's our answer! Phew.