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Source: in Poncelet's porism, me

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2643 posts

yetti

#1 h Jun 13, 2008, 4:32 PM • 4 Y

Y by Davi-8191, houkai, GuvercinciHoca, Adventure10

Let $\mathbf P = \left\{\triangle ABC\ |\ (I), (O) = \text{const}\right\}$ be a set of triangles with a common incircle $(I)$ and circumcircle $(O)$ . Let $X$ be arbitrary fixed point inside or on the incircle and let $\triangle DEF$ be the pedal triangle of $X$ with respect to a $\triangle ABC \in \mathbf P,$ with $D \in BC, E \in CA, F \in AB.$ Show that the sum $XD + XE + XF$ does not depend on the $\triangle ABC \in \mathbf P.$

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2643 posts

yetti

#2 h Jun 20, 2008, 6:25 PM • 1 Y

Y by Adventure10

1. When $X$ is no longer inside of the incircle, $\overline{XD} + \overline{XE} + \overline {XF} = \text{const}$ holds for all $\triangle ABC \in \mathbf P$ only if the distances for each triangle in $\mathbf P$ are properly directed - one or two are negative outside of the triangle.

2. Locus of $X$ such that $\overline{XD} + \overline{XE} + \overline {XF} = 0$ ?

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6194 posts

#3 h Jun 20, 2008, 6:54 PM • 3 Y

Y by Mindstormer, Adventure10, Mango247

Euler's theorem for pedal triangle states that

[DEF]/[ABC]=|R^2-OP^2|/4R^2

Note OP^2-R^2 is the power of P with respect to the circumcircle.

I'm not sure if this helps or not.

The main idea of the proof is extending AP, BP, CP to hit the circumcircle again at X,Y,Z, then XYZ ~ DEF and a bunch of other similarities...

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4146 posts

#4 h Mar 23, 2011, 9:32 PM • 3 Y

Y by baopbc, Adventure10, Mango247

Lemma. The locus of points $P,$ whose sum of oriented distances to the sidelines of $\triangle ABC$ is constant, is a perpendicular to the line connecting its incenter $I$ and circumcenter $O.$

Proof. Let $(\alpha:\beta:\gamma)$ be the trilinear coordinates of $P$ with respect to $\triangle ABC.$ Assume that the sum of the oriented distances from $P$ to $BC,CA,AB$ is a constant $k.$ Thus, the trilinear equation of its locus $\ell$ is given by

$\ell \equiv \frac{\alpha \cdot S}{a\alpha+b\beta+c\gamma}+\frac{\beta \cdot S}{a\alpha+b\beta+c\gamma}+\frac{\gamma \cdot S}{a\alpha+b\beta+c\gamma}=k \ , \ S=2|\triangle ABC|$

$\ell \equiv (S-ak)\alpha+(S-bk)\beta+(S-ck)\gamma=0$

Infinite point of $\ell$ for any $k$ is indeed $X_{513} \equiv (b-c:c-a:a-b) \Longrightarrow \ell \perp IO.$
_______________________________

Now, back to the problem, let $U$ be the fixed orthogonal projection of $X$ on $IO.$ Let $D',E,F'$ denote the orthogonal projections of $U$ on $BC,CA,AB.$ $M,N,L$ are the midpoints of $BC,CA,AB$ and $(I)$ touches $BC$ at $U.$ From the right trapezoid $IOMK$ with bases $OM \parallel IK,$ we get

$\overline{UD'}=\frac{\overline{OM} \cdot \overline{UI}+\overline{IK} \cdot \overline{UO}}{IO}=\frac{\overline{OM} \cdot \overline{UI}+r \cdot \overline{UO}}{IO}$

Similarly, we'll have the expressions

$\overline{UE'}=\frac{\overline{ON} \cdot \overline{UI}+r \cdot \overline{UO}}{IO} \ \ , \ \ \overline{UF'}= \frac{\overline{OL} \cdot \overline{UI}+r \cdot \overline{UO}}{IO}$

$\Longrightarrow \overline{UD'}+\overline{UE'}+\overline{UF'}= \frac{ (\overline{OM}+\overline{ON}+\overline{OL}) \cdot \overline{UI}+3r \cdot \overline{UO}}{IO}$

But, by Carnot's theorem we have $\overline{OM}+\overline{ON}+\overline{OL}=R+r$

$\Longrightarrow \overline{UD'}+\overline{UE'}+\overline{UF'}= \frac{ (R+r) \cdot \overline{UI}+3r \cdot \overline{UO}}{IO}=\text{const}$

Now, according to the previous lemma, we deduce that

$\overline{XD}+\overline{XE}+\overline{XF}=\overline{UD'}+\overline{UE'}+\overline{UF'}=\text{const}.$

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4146 posts

#5 h Jan 2, 2015, 6:54 PM • 4 Y

Y by baopbc, Kanep, GuvercinciHoca, Adventure10

Here is a synthetic proof to the lemma I referred in the previous post.

If we take points $Y,Z$ on rays $\overrightarrow{CA},\overrightarrow{BA},$ such that $CY=BZ=BC,$ then $OI \perp YZ$ (this has been discussed many times before). If $\delta_A,\delta_B,\delta_C$ denote the oriented distances from $P$ to $BC,CA,AB,$ then using oriented areas, we get

$[BZYC]=[PCB]+[PBZ]+[PZY]+[PYC]=\frac{BC \cdot (\delta_A+\delta_B+\delta_C)}{2}+[PZY].$

Hence, if $\delta_A+\delta_C+\delta_C$ is constant, then $[PZY]$ is constant $\Longrightarrow$ oriented distance from $P$ to $YZ$ is constant $\Longrightarrow$ $P$ moves on a line parallel to $YZ,$ i.e. perpendicular to $OI.$

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andria

#7 h Oct 30, 2015, 1:14 PM • 5 Y

Y by r3mark, Wizard_32, GuvercinciHoca, Adventure10, Mango247

Luis González wrote:

Lemma. The locus of points $P,$ whose sum of oriented distances to the sidelines of $\triangle ABC$ is constant, is a perpendicular to the line connecting its incenter $I$ and circumcenter $O.$

My proof:
Let $f(P)$ be the sum of distances of $P$ from $AB,AC,BC$ .
Let $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ be the vectors such that $|u|=|v|=|w|=1$ and $\overrightarrow{u}. \overrightarrow{BC}=0,\overrightarrow{v}. \overrightarrow{AC}=0,\overrightarrow{w}. \overrightarrow{AB}=0$ (see figure). Consider two arbitrary points $P,Q$ such that $f(P)=f(Q)$ . Let $X,Y,Z$ be the projections of $P$ on $BC,CA,AB$ respectively and $X',Y',Z'$ be the projections of $Q$ on $BC,CA,AB$ respectively. Note that $\sum\overline{PX}=\sum\overline{QX'}$ . Observe that:
$PX=\overrightarrow{PB}.\overrightarrow{u}$
$PY=\overrightarrow{PC}.\overrightarrow{v}$
$PZ=\overrightarrow{PA}.\overrightarrow{w}$
Hence $f(P)=\sum\overrightarrow{PB}.\overrightarrow{u}$ similarly $f(Q)=\sum\overrightarrow{QB}.\overrightarrow{u}$ so $0=f(P)-f(Q)=\overrightarrow{PQ}.(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w})\Longrightarrow PQ\perp \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}$ .
Let $M,N,P$ be the midpoints of arcs $BC,AC,AB$ respectively and $G$ be the centroid of $\triangle MNP$ clearly $I$ is orthocenter of $\triangle MNP$ . Hence $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\frac{1}{R}(\overrightarrow{OM}+\overrightarrow{ON}+\overrightarrow{OP})=3\overrightarrow{OG}=\overrightarrow{OI}\Longrightarrow PQ\perp OI$ .
Q.E.D
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824 posts

andria

#8 h Oct 30, 2015, 1:25 PM • 2 Y

Y by Adventure10, Mango247

Generalization of above problem:
Let $n\ge 3$ be a positive integer and $A_1A_2\dots A_n$ be an arbitrary $n-gon$ . then the locus of point $P$ such that $\sum\text{dis}(P,A_iA_{i+1})$ is constat is a line perpendicular to $\overrightarrow{u_1}+\overrightarrow{u_2}+\dots +\overrightarrow{u_n}$ where $|u_1|=|u_2|=\dots=|u_n|=1$ and $\overrightarrow{u_i}.\overrightarrow{A_iA_{i+1}}=0$ .
The solution in post #7# also works for this problem.

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