USA GEO 2003

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dreammath

82 posts

dreammath

#1 h Feb 16, 2004, 5:49 AM • 3 Y

Y by Pure_IQ, Adventure10, Mango247

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
$[PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]$
if and only if $P$ lies on at least one of the medians of triangle $ABC$ . (Here $[XYZ]$ denotes the area of triangle $XYZ$ .)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

StRyKeR

123 posts

StRyKeR

#2 h Feb 23, 2004, 11:50 PM • 2 Y

Y by Adventure10, Mango247

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.)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Philip_Leszczynski

327 posts

Philip_Leszczynski

#3 h Nov 11, 2005, 12:08 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Click to reveal hidden text

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Virgil Nicula

7054 posts

Virgil Nicula

#4 h Nov 11, 2005, 11:06 AM • 1 Y

Y by Adventure10

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. $\frac{(x_1+y_1)+y_2}{(x_2+z_2)+z_1}\cdot \frac{(y_1+z_1)+z_2}{(x_2+y_2)+x_1}\cdot \frac{(z_1+x_1)+x_2}{(y_2+z_2)+y_1}=1\Longleftrightarrow$

$\frac{1+(y_2-z_1)}{1-(y_2-z_1)}\cdot \frac{1+(z_2-x_1)}{1-(z_2-x_1)}\cdot \frac{1+(x_2-y_1)}{1-(x_2-y_1)}=1\Longleftrightarrow$ $(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longleftrightarrow$ $xyz=0\Longleftrightarrow$

$y_2=z_1\ \vee\ z_2=x_1\ \vee\ x_2=y_1\Longleftrightarrow$ $EF\parallel BC\ \vee\ FD\parallel CA\ \vee\ DE\parallel AB\Longleftrightarrow$ $P\in [AM]\cup [BN]\cup [CP]\ .$

This post has been edited 6 times. Last edited by Virgil Nicula, Mar 16, 2017, 7:19 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dieuhuynh

266 posts

dieuhuynh

#5 h May 12, 2006, 11:38 AM • 2 Y

Y by Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

who

190 posts

who

#6 h Jun 19, 2006, 11:20 AM • 2 Y

Y by Adventure10, Mango247

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

,my sol. is similar.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

probability1.01

2743 posts

probability1.01

#7 h Mar 4, 2007, 3:34 AM • 1 Y

Y by Adventure10

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Wolstenholme

543 posts

Wolstenholme

#8 h Aug 20, 2014, 3:28 AM • 6 Y

Y by Hydrogen-Helium, ThisIsASentence, Wizard_32, dchenmathcounts, Tafi_ak, Adventure10

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

va2010

1276 posts

va2010

#9 h Sep 15, 2014, 1:14 AM • 2 Y

Y by Adventure10, Mango247

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?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

EulerMacaroni

851 posts

EulerMacaroni

#10 h Jan 12, 2015, 5:24 PM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Wave-Particle

3690 posts

Wave-Particle

#11 h Mar 16, 2017, 6:58 AM • 2 Y

Y by Adventure10, Mango247

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

FISHMJ25

293 posts

FISHMJ25

#12 h Apr 3, 2018, 8:00 PM • 2 Y

Y by Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

purplish

299 posts

purplish

#13 h Apr 3, 2018, 8:49 PM • 2 Y

Y by Adventure10, Mango247

ratiborpaljevic33 wrote:

What is formula for surfice of triangle in bary

See http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/bary.pdf

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dchenmathcounts

2443 posts

dchenmathcounts

#15 h Jul 21, 2020, 1:35 AM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rafaello

1079 posts

rafaello

#16 h Jul 29, 2021, 11:10 PM • 1 Y

Y by MS_asdfgzxcvb

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Tafi_ak

309 posts

Tafi_ak

#17 h Dec 22, 2021, 1:32 PM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mogmog8

1080 posts

Mogmog8

#18 h Feb 5, 2022, 4:00 AM • 2 Y

Y by centslordm, megarnie

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. $\square$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

john0512

4175 posts

john0512

#19 h Jan 9, 2023, 10:25 PM

Y by

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.

This post has been edited 1 time. Last edited by john0512, Jan 9, 2023, 11:08 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Taco12

1757 posts

Taco12

#20 h Jan 25, 2023, 11:05 PM • 2 Y

Y by Mango247, Mango247

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.