Seven Triangles

Let the areas of $\triangle GPC$ , $\triangle APG$ , and $\triangle CPF$ be $a$ , $b$ , and $c$ respectively. By Ceva's theorem we have:

$\begin{aligned} \frac {CG}{GA}\cdot \frac {AE}{EB} \cdot \frac {BF}{FC} & = 1 \\ \implies \frac ab \cdot \frac {40}{72} \cdot \frac {84}c & = 1 \end{aligned}$

By Menelaus' theorem ,

$\begin{aligned} \frac {CG}{GA}\cdot \frac {AB}{BE} \cdot \frac {EP}{PC} & = 1 \\ \implies \frac ab \cdot \frac {112}{72} \cdot \frac {72}{c+84} & = 1 \end{aligned}$

Therefore,

$\begin{aligned} \frac {112}{72} \cdot \frac {72}{c+84} & = \frac {40}{72} \cdot \frac {84}c \\ \implies \frac {c+84}{12} & = \frac c5 \\ c & = 60 \end{aligned}$

From $\dfrac ab \cdot \dfrac {40}{72} \cdot \dfrac {84}c = 1 \implies b = \dfrac {40}{72} \cdot \dfrac {84}{60} a = \dfrac 79 a$ . Note that the ratio of areas:

$\begin{aligned} \frac {[ACE]}{[BCE]} & = \frac {[APE]}{[BPE]} \\ \frac {a+b+40}{c+84+72} & = \frac {40}{72} \\ \frac {16}9 a & = \frac {40}{72}\cdot 216 - 40 = 80 \\ \implies a & = \boxed{45} \end{aligned}$

Let's solve using Mass Point Geometry

Now, let's balance $AB$ about $E$

Now, $\frac{AE}{EB} = \frac{\Delta APE}{\Delta BPE} = \frac{40}{72}$

So, let's keep $9m$ mass at $A$ and $5m$ mass at $B$ and $AB$ is balanced about $E$ . So, mass at $E = 5m + 9m = 14m$

Let's balance $AF$ about concurrency point $P$

$\frac{AP}{PF} = \frac{\Delta APB}{\Delta BPF} = \frac{112}{84}$

Now, as there is $9m$ mass on $A$ , so there is $12m$ mass at $F$ giving mass at $P$ as $9m + 12m =21m$ .

Now, you can find proof on wiki why all sides will be balanced about respective foot of cevians and all cevians are balanced about $P$

So, mass at $C$ is $12m - 5m = 7m$ and mass at $G$ is $7m + 9m = 16m$

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Now, as $BC$ is balanced about $F$ , so

$(5m)(\Delta BPF) = (7m)(\Delta CPF)$

$5(84) = 7(\Delta CPF)$ $\Delta CPF = 60$

Now, as $BG$ is balanced about $P$ , so

[(5m)(\Delta BPC) = (16m)(\Delta GPC)]

$5(144) = 16(\Delta GPC)$ $\color{#3D99F6}{\boxed{\Delta GPC = 45}}$

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Solution may seem long but it is actually very short and time saving while solving.

@Pi Han Goh, I see this type of problem has some interesting history on Brilliant. I was unaware of the Ladder Theorem. My inspiration for the problem came from this paper , which seems to have rediscovered the Ladder Theorem. In the second equation of Theorem 2.1, the author writes:

$K_5 = \dfrac{K_1K_2K_3^2(K_1+K_2)}{(K_2(K_1+K_2)-K_1K_3)(K_2(K_1+K_2+K_3)-K_1K_3)}$

or in your nomenclature:

$b = \dfrac{fed^2(e+f)}{(e(f+e)-fd)(e(f+e+d)-fd)}$

But solving the three linear equations you quote from the Ladder Theorem, we get the same equation , with only minor rearrangement of terms. Plugging in $d=84, e=72, f=40$ , we get:

$b = \dfrac{40\cdot72\cdot84^2\cdot(72+40)}{(72\cdot(40+72)-40\cdot84)(72\cdot(40+72+84)-40\cdot84)} = \boxed{45}$

x is area above y (?) and z below y, solve for y using the following relations

(84)(40)y=(x*z)(72),

84(x+y+z)=(40+72+84)x,

[(40+72+z)/(84+x+y)]=(z/y)

Answer

x=60, z=35

y=45

Let $X$ be the area of $\triangle AGP$ , $Y$ be the area of $\triangle GPC$ , and $Z$ be the area of $\triangle CFP$ :

If $h_1$ is the height of $\triangle AEP$ and $\triangle EPB$ and $h_2$ is the height of $\triangle AEC$ and $\triangle ECB$ , then $\cfrac{AE}{EB} = \cfrac{\frac{1}{2} \cdot h_1 \cdot AE}{\frac{1}{2} \cdot h_1 \cdot EB} = \cfrac{A_{\triangle AEP}}{A_{\triangle EPB}} = \cfrac{40}{72}$ and $\cfrac{AE}{EB} = \cfrac{\frac{1}{2} \cdot h_2 \cdot AE}{\frac{1}{2} \cdot h_2 \cdot EB} = \cfrac{A_{\triangle AEC}}{A_{\triangle ECB}} = \cfrac{X + Y + 40}{Z + 84 + 72}$ , so that $\cfrac{40}{72} = \cfrac{X + Y + 40}{Z + 84 + 72}$ .

By similar arguments, $\cfrac{Z}{84} = \cfrac{X + Y + Z}{40 + 72 + 84}$ and $\cfrac{Y}{X} = \cfrac{Y + Z + 84}{X + 40 + 72}$ .

These three equations solve to $X = 35$ , $Z = 60$ , and $Y = \boxed{45}$ .

Let $(a,b,c,d,e,f)$ denote the areas of these 6 triangles as shown. From Mark Henning's comment under Michale Mendrin's solution We are left to solve a system of linear equations, $\begin{cases}(a + b)d = (e + f) c \\ (c+d)f = (a+b) e \\ (c+ d) a = (e+f)b \end{cases} \quad \implies \quad (a,b,c) = (35,\boxed{45},60).$