I'm not seeing how this is an inequality problem, it should be moved to the geometry forums A very heavyily algebraic solution: Let $a = [PDC], b = [PAE], c = [BFP]$. Then: $1/a = (2 + c)/(1 + b + a) \implies 1 + b = a + ac$ $1/c = (2+b)/(1+c+a) \implies 1 + a = c + bc$ $1/b = (2+a)/(1+b+c) \implies 1 + c = b + ab$ $a = \frac{b+1}{c+1}$ $b = \frac{c+1}{a+1}$ $c = \frac{a+1}{b+1}$ $b = \frac{c+1}{\frac{b+1}{c+1}+1} = \frac{(c+1)^2}{b+c+2}$ $b^2 + b(2+c) - (c+1)^2 = 0$ $b = \frac{-2-c \pm \sqrt{(c+2)^2 + 4(c+1)^2}}{2} = \frac{-2-c + \sqrt{5c^2 + 12c + 8}}{2}$ $b+1 = \frac{-c + \sqrt{5c^2 + 12c + 8}}{2}$ Hence $a = \frac{-c + \sqrt{5c^2 + 12c + 8}}{2c + 2}$ $a+1 = \frac{c+2 + \sqrt{5c^2 + 12c + 8}}{2c + 2}$ $c = \frac{\frac{c+2 + \sqrt{5c^2 + 12c + 8}}{2c + 2}}{\frac{-c + \sqrt{5c^2 + 12c + 8}}{2}}$ $c^2 + c = \frac{c+2 + \sqrt{5c^2 + 12c + 8}}{-c + \sqrt{5c^2 + 12c + 8}}$ $c^2 + c = \frac{(c+2 + \sqrt{5c^2 + 12c + 8})(c + \sqrt{5c^2 + 12c + 8})}{4c^2 + 12c + 8}$ $4c^4 + 16c^3 + 20c^2 + 8c = 6c^2 + 14c + 8 + (2c+2)\sqrt{5c^2 + 12c + 8}$ $4c^4 + 16c^3 + 14c^2 - 6c - 8 = (2c+2)\sqrt{5c^2 + 12c + 8}$ $2c^3 + 6c^2 + c - 4 = \sqrt{5c^2 + 12c + 8}$ $(2c^3 + 6c^2 + c - 4)^2 - 5c^2 - 12c - 8 = 0$ $4(c-1)(c+1)(c+2)(c^3 + 4c^2 - 3c - 1) = 0$ Observe that $c=1$ is the only positive root except a root around $0.25$ of the cubic. You can bound the root between $0.2$ and $0.25$. But for that root $2c^3 + 6c^2 + c - 4 < 0$ and hence it is not actually a solution for $c$. Hence the only solution is $c=1$, so $a=b=1$ as well and therefore $[ABC] = 6$.

No need to do all that, lol.. Just use Ceva's Theorem to get $abc = 1.$, and then it's a simple application of basic inequalities with the equations you found..

Yes that's true, the solution I had above was very dumb but straightforward, it only took about 10~15 minutes to bash out actually.

Hmm actually you don't need inequalities.. Summing up your equations you get $ab + bc +ca = 3$, and with $abc = 1$, we have : \[b+b^2 = ab + 1\]\[c+ c^2 = bc+1\]\[a +a^2 = ac + 1\] Summing again, we get $a+b+c + (a+b+c)^2 - 12 = 0 \implies a+b+c=3.$ How did you factor that huge polynomial and get the root estimations of your cubic so quickly without a calculator?

Well $c=1$ is obviously a root. Playing around a little I found $-1,-2$ are also roots so I factored it then, so it sufficed to consider the cubic term. The cubic has three real roots (easy to verify) and there is only one positive root which happens to be between 0 and 1. Then guessing I found the root was less than $0.5$, then less than $0.25$ which was enough to establish the root thing was negative yielding $c=1$ as the only actual solution.

From $ab+bc+ac = 3$ and $abc = 1$ you can say $a=b=c=1$ because of $AM - GM$.. I think this was a good problem 1, not completely straightforward but very natural if you know how to prove ceva for example.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=55209. It is posted in mathlinks more than 6 years ago.

how do you get ab+bc+ca=3 ??

Here is a more general problem. $P$ is a point inside triangle $ABC$, and $AP, BP,CP$ meet the opposite sides at $D,E,F$ respectively. If the areas of three of the six triangles so formed are equal then $P$ is the centroid of triangle $ABC$.

APMO 2012 P1 wrote: Let $ P $ be a point in the interior of a triangle $ ABC $, and let $ D, E, F $ be the point of intersection of the line $ AP $ and the side $ BC $ of the triangle, of the line $ BP $ and the side $ CA $, and of the line $ CP $ and the side $ AB $, respectively. Prove that the area of the triangle $ ABC $ must be $ 6 $ if the area of each of the triangles $ PFA, PDB $ and $ PEC $ is $ 1 $. Maybe Bary Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$,let $P=(p,q,r)$ Its easy to see that $D=\left(0,\dfrac{q}{q+r},\dfrac{r}{q+r}\right),E=\left(\dfrac{p}{p+r},0,\dfrac{r}{r+p}\right),F=\left(\dfrac{p}{p+q},\dfrac{q}{p+q},0\right)$.The area of the respective triangles are - $X=\left\{\Delta{PAE}=\dfrac{qr}{r+p},\Delta{PEC}=\dfrac{pq}{r+p}\right\},Y=\left\{\Delta{PCD}=\dfrac{pq}{q+r},\Delta{PDB}=\dfrac{rp}{q+r}\right\},Z=\left\{\Delta{PBF}=\dfrac{rp}{p+q},\Delta{PFA}=\dfrac{qr}{p+q}\right\}$ Observe that the area of any group $\in \{p,q,r\}$.Now of any 3 selected triangles 2 must belong to one of the group $\left(X,Y,Z\right)$.So we have to solve the equation $\dfrac{pq}{q+r}=\dfrac{qr}{r+p}=\dfrac{rp}{p+q}$ .I feel sleepy so I will continue tommorow anyone wanting to modify my work is welcomed. $\mathbf{NOTE}$-I originally had created the sets $(X,Y,Z)$ to solve the problem proposed by @synthetic_potato but.....

Quite hard for a #1 (especially if you are anti-bash) First, we make some synthetic observation. Since $[PFA]=[PDB]$, we have $$\frac{AP}{PD} = \frac{\mathrm{dist}(B,AD)}{\mathrm{dist}(F,AD)} = \frac{AB}{AF}$$and cyclic relation holds. Now we resort to barycentric coordinates w.r.t. $\triangle ABC$. Set $P=(a:b:c)$. Then we clearly have $AP : PD = b+c:a$ and $AB : AF = a+b : b$. Thus $$\frac{b+c}{a} = \frac{a+b}{b}\implies a(a+b) = b(b+c).$$Cyclic relations hold so $a(a+b)=b(b+c)=c(c+a)$. To solve this system of equations, assume that $a>b$, then $a+b<b+c$ $\implies a<c$, which in turn implies $a+b>c+a$ $\implies b>c$, contradiction. Similarly $a<b$ causes contradiction hence $a=b=c$. This means $P$ is a centroid and the result became obvious.

Let $[PCD] = x, [PAE] = y, [PBF] = z$. By Ceva's theorem, $xyz = 1$. Also $\frac{CP}{PF} = \frac{x+1}{z} = y+1$, so $x+1 = yz + z$ (similarly, $y+1 = zx + x$ and $z+1 = xy+y$). If $x = 1$, then we can solve to get $x=y=z=1$. Otherwise, WLOG $x > 1$ and $y, z < 1$, or $x < 1$ and $y, z > 1$. In the first case, $2 < x+1 = yz+z < 2$, and in the second case, $2 > x + 1 = yz + z > 2$, contradiction.

Really Nice problem Let [PFB] = x, [PDC] = y and [PEA] = z. Claim1 : xyz = 1. Proof : By Ceva Theorem we have 1/x . 1/y . 1/z = 1 so xyz = 1. Claim2 : xy + yz + zx = 3. Proof : Note that [APB]/[CPB] = [AEP]/[CEP] so x+1/y+1 = z so x+1 = zy + z with same approach we have y+1 = xz + x and z+1 = yx + y so xy + yz + zx = 3. Now with AM - GM we have xy = yz = zx so x = y = z = 1 so [ABC] = 6. we're Done.

Let $x=[\triangle APE], y=[\triangle BPF], z=[\triangle CPD]$. By Ceva: $xyz=1$. Also $\frac{[\triangle CPE]}{[\triangle CPB]}=\frac{[\triangle APE]}{[\triangle APB]}$, or $1+y=x+xz=x+\frac1y$ or $(y+1)\cdot \frac{y-1}y=x-1$. If one of $x,y,z$ is one, then the others are clearly one. Otherwise, by multiplying cyclically the last equation, we get $(x+1)(y+1)(z+1)=1$, which is impossible.

Lemma: If $x,y,z$ are positive real numbers such that $$xyz=1, xy+xz+yz=3,$$then $$x=y=z=1.$$Divide the second equation by $xyz$ so that $$1/x+1/y+1/z=3.$$Let $a=1/x,b=1/y,c=1/z.$ Then, this becomes $$a+b+c=3,abc=1.$$By AM-GM, $$3=a+b+c\geq 3\sqrt[3]{abc}=3,$$so we must have $a=b=c$ and thus $x=y=z=1.$ Let $$[APE]=x,[BPF]=y,[CPD]=z.$$Then, $xyz=1$ by the fact that $x=AE/CE$ etc. and Ceva. Furthermore, $$\frac{[ABD]}{[ACD]}=\frac{BD}{CD}=\frac{[BPD]}{[CPD]}.$$Hence, $$\frac{y+2}{x+z+1}=\frac{1}{z}.$$This rearranges to $$yz=x-z+1.$$Similarly, $$zx=y-x+1,xy=z-y+1.$$Adding these three equations, we get that $$xy+xz+yz=3.$$By our lemma, it follows that $x=y=z=1$, so we are done.

Denote $[BPF] = a, [PDC] = b, [PEA] = c$. Observe from Ceva's theorem that $abc = 1$. Observe that $$b = \frac{CD}{DB} = \frac{b+c+1}{a+2} \implies b + c + 1 = ba + 2b.$$Similarly, $$a+b+1 = ac + 2a$$and $$c + a + 1 = cb + 2c.$$Adding up, we see that $$ab + bc + ca = 3.$$But now $\frac{ab + bc + ca}{3} = \frac{3}{3} = 1 = 1^{\frac{2}{3}} = (abc) ^{\frac{2}{3}}$, and thus the equality case of AM-GM holds. Therefore $ab = bc = ca = 1 \implies a = b = c = 1$. Therefore the area of $ABC$ is $a + b + c + 3 = 6$. $\square$

We proceed by barycentric coordinates. Let $P = (a, b, c)$. Then we have cyclically that $\frac{[AFP]}{[ABC]} = \frac{bc}{a+b}$. Thus cyclically, we have $\frac{bc}{a+b} = \frac{ca}{b+c}$ so $b(b+c) = a(a+b)$ so $b - ab = a - ac$ so $b = a(1+b-c)$. Now multiplying the final equation cyclically, we have $\prod_{\textrm{cyc}}(1+b-c) = 1$, so by AM-GM it follows that $a = b = c$. Thus $P$ is the centroid, so it follows immediately that $[ABC] = 6$.