Let BD / DC = a, CE / EA = b, and AF / FB = c. First, by Ceva's theorem, abc = 1. Let [APF] = s_1, [BPF] = s_2, [BPD] = s_3, [CPD] = s_4, [CPE] = s_5, and [APE] = s_6, and [ABC] = S. 1/(a + 1) = DC / (BD + DC) = DC / BC = (s_4 + s_5 + s_6) / S. 1/(b + 1) = (s_2 + s_1 + s_6) / S. 1/(c + 1) = (s_2 + s_3 + s_4) / S. Then \sum 1/(a+1) = (s_1 + s_2 + s_3 + s_4 + s_5 + s_6 + s_2 + s_4 + s_6) / S = 3/2, since s_1 + s_3 + s_5 = s_2 + s_4 + s_6 = S/2. Then multiplying (a+1)(b+1)(c+1) to both sides, 2(b+1)(c+1) + 2(a+1)(c+1) + 2(a+1)(b+1) = 3(a + 1)(b + 1)(c + 1) reduces to ab + bc + ca - a - b - c = 3 - 3abc = 0, since abc = 1. But then ab + bc + ca - a - b - c = ab + bc + ca - a - b - c + abc - 1 = 0 since abc - 1 = 0. Then (a - 1)(b - 1)(c - 1) = 0, from which the result follows. (One of a, b, c equals 1, which implies one of the ratios AF/FB, BD/DC, CE/EA equals 1 and hence P lies on a median.)

PP. Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Prove that $[PAF]+[PBD]+[PCE]=\frac S2\iff P$ lies on at least one of the medians of triangle $ABC$ . Proof. Denote the area $[XYZ]$ of the triangle $\triangle XYZ$ . I will suppose w.l.o.g. that $[ABC]=2$ and I note: $\bullet\ M,N,P$ - the midpoints of the sides $[BC],[CA],[AB]$ respectively; $\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$ $[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$ $\bullet\ x=y_2-z_1\ ,\ y=z_2-x_1\ ,\ z=x_2-y_1\ .$ Thus, $\boxed{x_1+y_1+z_1=x_2+y_2+z_2=1\ ;\ x+y+z=0}\ .$ From the Ceva's theorem results the relation $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1$ , i.e. $\frac{(x_1+y_1)+y_2}{(x_2+z_2)+z_1}\cdot \frac{(y_1+z_1)+z_2}{(x_2+y_2)+x_1}\cdot \frac{(z_1+x_1)+x_2}{(y_2+z_2)+y_1}=1\Longleftrightarrow$ $\frac{1+(y_2-z_1)}{1-(y_2-z_1)}\cdot \frac{1+(z_2-x_1)}{1-(z_2-x_1)}\cdot \frac{1+(x_2-y_1)}{1-(x_2-y_1)}=1\Longleftrightarrow$ $(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longleftrightarrow $ $xyz=0\Longleftrightarrow$ $y_2=z_1\ \vee\ z_2=x_1\ \vee\ x_2=y_1\Longleftrightarrow $ $EF\parallel BC\ \vee\ FD\parallel CA\ \vee\ DE\parallel AB\Longleftrightarrow$ $P\in [AM]\cup [BN]\cup [CP]\ .$

it is the same my solution, I prove that:abc=xyz,a+b+c=x+y+z and ab+bc+ca=xy+yz+zx

,due to tishio semiya,japan and was published as crux pro. bet 95-00. ,my sol. is similar.

Let a, b, and c denote the respective areas of PAB, PBC, and PCA. We then have [PAF] = ac/(b+c) and analogous, so our task is to determine for what a, b, and c we have ac/(b+c) + ba/(a+c) + bc/(a+b) = (a+b+c)/2. It would suffice to show that this equation implies either a = b, b = c, or c = a. Indeed, suppose for sake of contradiction that none of these equalities were true for some (a', b', c') satisfying the equation. Then expand out the equation and consider it as a cubic in a. By inspection, the cubic evaluates at zero when a = b, a = c, and a = -b - c. Since this equation must also hold at a = a', the cubic has four roots and must be identically zero. However, plugging in a = 0 gives a contradiction, so the result is proven.

This question almost begs the use of barycentric coordinates. Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ P = (r : s : t) $ for some real $ r, s, t > 0 $. Then $ D = (0 : s : t), E = (r : 0 : t), F = (r : s : 0) $. Now $ [PAF] = \frac{1}{(r + s + t)(r + s)}\begin{pmatrix}1 & 0 & 0 \\ r & s & 0 \\ r & s & t \end{pmatrix}[ABC] = \frac{st}{(r + s + t)(r + s)}[ABC] $ so the problem condition is equivalent to $ \frac{r + s + t}{2} - \frac{st}{r + s} - \frac{tr}{s + t} - \frac{rs}{t + r} = 0 $ which upon expanding becomes $ \frac{(r - s)(s - t)(t - r)(r + s + t)}{2(r + s)(s + t)(t + r)} = 0 $ and so $ r, s, t $ are not all distinct which immediately implies the desired result.

Sorry to revive this, but the polynomial given $(r-s)(s-t)(t-r)(r+s+t)$ was 2014 USAMO #3. Is there something similar between these two problems?

va2010 wrote: Sorry to revive this, but the polynomial given $(r-s)(s-t)(t-r)(r+s+t)$ was 2014 USAMO #3. Is there something similar between these two problems? Not really. Both of these problems utilize the shoelace determinant, but to different ends, USAMO 2014 #3 uses it to construct a function which gives a shoelace determinant of 0 (making the points collinear) iff each of the values are distinct, which clearly can be seen by the determinant. This problem WANTS us to show that atleast two of $x,y,z$ are equal, because this would imply that the areas of these two triangles are equal, which clearly occurs with the median. So, to answer your original question, no, but kind of.

Wolstenholme wrote: This question almost begs the use of barycentric coordinates. Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ P = (r : s : t) $ for some real $ r, s, t > 0 $. Then $ D = (0 : s : t), E = (r : 0 : t), F = (r : s : 0) $. Now $ [PAF] = \frac{1}{(r + s + t)(r + s)}\begin{pmatrix}1 & 0 & 0 \\ r & s & 0 \\ r & s & t \end{pmatrix}[ABC] = \frac{st}{(r + s + t)(r + s)}[ABC] $ so the problem condition is equivalent to $ \frac{r + s + t}{2} - \frac{st}{r + s} - \frac{tr}{s + t} - \frac{rs}{t + r} = 0 $ which upon expanding becomes $ \frac{(r - s)(s - t)(t - r)(r + s + t)}{2(r + s)(s + t)(t + r)} = 0 $ and so $ r, s, t $ are not all distinct which immediately implies the desired result. What is formula for surfice of triangle in bary

ratiborpaljevic33 wrote: What is formula for surfice of triangle in bary See http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/bary.pdf

Say $P$ has barycentrics $\left(\dfrac{1}{a}:\dfrac{1}{b}:\dfrac{1}{c}\right).$ Then we can assign masses of $a,b,c$ to $A,B,C.$ Notice that the condition $P$ lies on at least one median of $\triangle ABC$ is the same as the condition that at least two of $a,b,c$ are equal. Without loss of generality, say $[ABC]=1.$ Note that by mass points, $BD=BC\frac{c}{b+c}$ and the height of the altitude from $P$ to $BD$ is $h_a\frac{a}{a+b+c},$ where $h_a$ is the height of the $A$-altitude. Thus $[PBD]=\frac{BC\cdot h_a}{2}\cdot\frac{c}{b+c}\cdot\frac{a}{a+b+c}=\frac{ca}{(b+c)(a+b+c)}.$ Analogously, \[[PCE]=\frac{ab}{(c+a)(a+b+c)}\]\[[PAF]=\frac{bc}{(a+b)(a+b+c)}.\]Now consider the equation \[\frac{ca}{(b+c)(a+b+c)}+\frac{ab}{(c+a)(a+b+c)}+\frac{bc}{(a+b)(a+b+c)}=\frac{1}{2}.\]We can multiply by $(a+b+c)(b+c)(c+a)(a+b)$ and instead consider the polynomial \[ca(c+a)(a+b)+ab(a+b)(b+c)+bc(b+c)(c+a)-\frac{1}{2}(a+b+c)(a+b)(b+c)(c+a).\]Note that no solutions will be lost from this. We claim that $a-b$ and $a+b+c$ are roots of this polynomial. To verify the former, let $a=b$ and note that \[ca(c+a)(2a)+a^2(2a)(c+a)+ca(c+a)^2-\frac{1}{2}(2a+c)(2a)(c+a)^2=\]\[(c+a)(2ca^2+2a^3+c^2a+ca^2-a(c+a)(2a+c))=0.\]By symmetry, $b-c$ and $c-a$ are roots. This verifies the if direction. To verify the latter, let $a+b+c=0$ and note that \[ca(-b)(-c)+ab(-c)(-a)+bc(-a)(-b)=\]\[abc(a+b+c)=0.\] Note there cannot be any other roots as our polynomial is quartic, and we have found four roots of degree one. In the context of the problem, $a+b+c=0$ is an extraneous root. This verifies the only if direction.

Let $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $P=(x,y,z)$. Then, $D=(0:y:z)$, $E=(x:0:z)$ and $F=(x:y:0)$. Thus, we need to show that $$\frac{yz}{x+y}+\frac{xz}{y+z}+\frac{xz}{x+z}=\frac{1}{2}$$is true iff two of $x,y,z$ are equal to each other with the constraint that $x+y+z=1$. Now, \begin{align*} \frac{yz}{x+y}+\frac{xz}{y+z}+\frac{xy}{x+z}&=\frac{1}{2}\Longleftrightarrow\\ \frac{y(1-x-y)}{x+y}+\frac{(1-y-z)z}{y+z}+\frac{x(1-x-z)}{x+z}&=\frac{1}{2}\Longleftrightarrow\\ \left(\frac{1}{x+y}-1\right)y+\left(\frac{1}{y+z}-1\right)z+\left(\frac{1}{x+z}-1\right)x&=\frac{1}{2}\Longleftrightarrow\\ \frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z}&=\frac{3}{2}\Longleftrightarrow\\ \frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z}&=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z}\Longleftrightarrow\\ \frac{x-z}{x+z}+\frac{y-x}{x+y}+\frac{z-y}{y+z}&=0\Longleftrightarrow\\ \frac{x-z}{x+z}+\frac{(y-x)(y+z)+(z-y)(x+y)}{(x+y)(y+z)}&=0\Longleftrightarrow\\ \frac{x-z}{x+z}+\frac{2yz-2xy}{(x+y)(y+z)}&=0\Longleftrightarrow\\ (x-z)\left(\frac{1}{x+z}-\frac{2y}{(x+y)(y+z)}\right)&=0\Longleftrightarrow\\ (x-z)\left(\frac{(x+y)(y+z)-2y(x+z)}{(x+y)(y+z)(x+z)}\right)&=0\Longleftrightarrow\\ (x-z)\left(\frac{y^2+xz-xy-yz)}{(x+y)(y+z)(x+z)}\right)&=0\Longleftrightarrow\\ \frac{(x-z)(x-y)(z-y)}{(x+y)(y+z)(x+z)}&=0. \end{align*}We are done.

Wolstenholme wrote: This question almost begs the use of barycentric coordinates. Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ P = (r : s : t) $ for some real $ r, s, t > 0 $. Then $ D = (0 : s : t), E = (r : 0 : t), F = (r : s : 0) $. Now $ [PAF] = \frac{1}{(r + s + t)(r + s)}\begin{pmatrix}1 & 0 & 0 \\ r & s & 0 \\ r & s & t \end{pmatrix}[ABC] = \frac{st}{(r + s + t)(r + s)}[ABC] $ so the problem condition is equivalent to $ \frac{r + s + t}{2} - \frac{st}{r + s} - \frac{tr}{s + t} - \frac{rs}{t + r} = 0 $ which upon expanding becomes $ \frac{(r - s)(s - t)(t - r)(r + s + t)}{2(r + s)(s + t)(t + r)} = 0 $ and so $ r, s, t $ are not all distinct which immediately implies the desired result. This is pretty nice. My approach is mostly similar to this.

Use barycentrics on $\triangle ABC.$ Let $P=(p,q,r)$ so $D=(0:q:r),E=(p:0:r),$ and $F=(p:q:0).$ Then, \begin{align*}\frac{[AFP]}{[ABC]}&=\frac{1}{p+q}\begin{vmatrix}1&0&0\\p&q&0\\p&q&r\end{vmatrix}=\frac{\begin{vmatrix}q&0\\q&r\end{vmatrix}}{p+q}=\frac{qr}{p+q}.\end{align*}Hence, the area condition is true if and only if $$0=\frac{qr}{p+q}+\frac{pr}{q+r}+\frac{pq}{p+r}-\frac{p+q+r}{2}=\frac{(p-q)(p-r)(r-q)(p+q+r)}{(p+q)(q+r)(r+p)}.$$This is equivalent to $p,q,$ and $r$ being non-distict. $\square$

ah Let $P=(x,y,z)$ in barycenric coordinates. We then have $$D=(0,y,z),E=(x,0,z),F=(x,y,0).$$Being careful that all areas are of the same sign, we have $$\frac{[AFP]+[BDP]+[CEP]}{[ABC]}=\sum_{cyc} \frac{1}{(x+y)(x+y+z)}det{(1,0,0;x,y,0;x,y,z)}$$$$=\sum_{cyc} \frac{yz}{(x+y)(x+y+z)}.$$We want to show that $$\frac{yz}{x+y}+\frac{zx}{y+z}+\frac{xy}{z+x}=\frac{1}{2}(x+y+z)$$if and only if either $x=y$, $y=z,$ or $z=x.$ After expanding, this becomes $$2(yz^3+zx^3+xy^3+sym(2,1,1)+cyc(2,2,0))=sym(2,1,1)+sym(3,1,0)+sym(2,2,0)+sym(2,1,1)$$Note that $sym(2,2,0)=2cyc(2,2,0)$, so this cancels to just $$2(yz^3+zx^3+xy^3)=sym(3,1,0)$$$$yz^3+zx^3+xy^3-(y^3z+z^3x+x^3y)=0.$$Finally, this factors as $$(x-y)(y-z)(z-x)(x+y+z)=0,$$and since $x+y+z=1$ this is true if and only if one two of $x,y,z$ are equal, which is equivalent to $P$ lying on a median.

writing this up for OTIS (ive solved this at least 3 times before but never wrote it up) Apply barycentric coordinates on $\triangle ABC$. Set $P=(l:m:n)$, so by parameterization we have $D=(0:m:n), E=(l:0:n), F=(l:m:0)$. Taking determinants and simplifying reduces the problem to $$\frac{(l-m)(l-n)(n-m)(l+m+n)}{(m+n)(l+m)(l+n)}=0$$which immediately implies the desired.