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Source: Problem 7 from CWMO 2007

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1363 posts

Erken

#1 h Nov 18, 2007, 9:32 AM • 2 Y

Y by Adventure10, Mango247

Let $P$ be an interior point of an acute angled triangle $ABC$ . The lines $AP,BP,CP$ meet $BC,CA,AB$ at points $D,E,F$ respectively. Given that triangle $\triangle DEF$ and $\triangle ABC$ are similar, prove that $P$ is the centroid of $\triangle ABC$ .

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2643 posts

yetti

#2 h Nov 19, 2007, 2:30 AM • 1 Y

Y by Adventure10

Let $\triangle A'B'C'$ be the pedal triangle of the $\triangle ABC$ WRT a point Q, with pedals $A' \in BC,\ B' \in CA,\ C' \in AB.$ Any $\triangle DEF$ with $D \in BC,\ E \in CA,\ F \in AB$ directly similar to the pedal $\triangle A'B'C'$ is obtained by a spiral similarity with center Q. If some other $\triangle DEF \sim \triangle A'B'C'$ existed, which could not be obtained in this way, one could pick the $\triangle D'E'F'$ obtained form the pedal $\triangle A'B'C'$ by the spiral similarity with center Q and equally inclined as $\triangle DEF.$ The $\triangle DEF \sim \triangle D'E'F$ would have parallel sides, they would have to be centrally similar with some similarity center, but this similarity center would have to be simultaneously identical with the vertices A, B, C of the original $\triangle ABC.$ If A', B', C' are midpoints of BC, CA, AB, the medial triangle $\triangle A'B'C'$ of the $\triangle ABC$ is directly similar to it and it is the pedal triangle WRT the circumcenter O. Since $\frac{A'B}{A'C} \cdot \frac{B'C}{B'A} \cdot \frac{C'A}{C'B} = 1,$ $AA', BB', CC'$ concur (at the centroid P). Any $\triangle DEF \sim \triangle ABC \sim \triangle A'B'C'$ with $D \in BC,\ E \in CA,\ F \in AB$ is obtained by a spiral similarity with center O. But then either $BD > BA',\ CE > CB',\ AF > AC'$ and $CD < CA',\ AE < AB',\ BF < BC'$ or the other way around, $\frac{DB}{DC} \cdot \frac{EC}{EA} \cdot \frac{FA}{FB} > 1$ or $< 1,$ and AD, BE, CF do not concur.

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7 posts

dlt5

#3 h Nov 19, 2007, 11:12 AM • 2 Y

Y by Adventure10, Mango247

This my solution:
We see $\triangle DEF \sim \triangle ABC$ and they are same directly
-> existed a transform propitious:
$D \mapsto A$
$E \mapsto B$
$F \mapsto C$
and the transform propitious only has one immovable: O
after that we'll prove that: $O\equiv P$

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1363 posts

Erken

#4 h Nov 19, 2007, 1:23 PM • 2 Y

Y by Adventure10, Mango247

Does anybody know the name and the proof of the following theorem:
Let $\triangle ABC$ and $\triangle A'B'C'$ be two similar triangles,such that $P\in AA'\cap BB'\cap CC'$ .Prove that $P$ is homothety center.

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6555 posts

#5 h Nov 19, 2007, 3:11 PM • 5 Y

Y by YanYau, chasse-neige, Adventure10, Mango247, and 1 other user

Erken wrote:

Does anybody know the name and the proof of the following theorem:
Let $\triangle ABC$ and $\triangle A'B'C'$ be two similar triangles,such that $P\in AA'\cap BB'\cap CC'$ .Prove that $P$ is homothety center.

It is called "Wrong Theorem".

Take two circles k and k' intersecting at two points P and Q, and three points A, B, C on k. Let the lines AP, BP, CP intersect k' again at A', B', C'. The triangles ABC and A'B'C' are directly similar, and P lies on the lines AA', BB', CC', but the triangles ABC and A'B'C' are only homothetic if the circles k and k' are tangent to each other.

darij

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1363 posts

Erken

#6 h Nov 19, 2007, 4:26 PM • 2 Y

Y by Adventure10, Mango247

Uuups

,maybe i misunderstood something because one guy solved it using this theorem,and he score max points for this problem....

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1648 posts

#7 h Nov 19, 2007, 6:38 PM • 2 Y

Y by Adventure10, Mango247

darij grinberg wrote:

It is called "Wrong Theorem".
darij

I was just ready to ask , if such a theorem is valid , but your post is much better !!!

Babis

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45 posts

#8 h Aug 14, 2013, 7:17 AM • 1 Y

Y by Adventure10

Erken wrote:

Let $P$ be an interior point of an acute angled triangle $ABC$ . The lines $AP,BP,CP$ meet $BC,CA,AB$ at points $D,E,F$ respectively. Given that triangle $\triangle DEF$ and $\triangle ABC$ are similar, prove that $P$ is the centroid of $\triangle ABC$ .

Just assume that $FE$ and $BC$ aren't parallel. Let $FE$ and $BC$ intersects in $A_1$ , simillary denote $B_1$ and $C_1$ . WLOG assume that $A_1$ lies on extension $CB$ behind $B$ . Then by simple angle chasing we get that $B_1$ lies on extension $AC$ behind $C$ and $C_1$ lies on extension $BA$ behind $A$ . But by Desargues theorem, we know that $A_1$ , $B_1$ and $C_1$ are collinear which is clearly impossible...

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290 posts

jred

#9 h Jul 18, 2014, 8:25 AM • 3 Y

Y by chasse-neige, Adventure10, Mango247

Erken wrote:

Does anybody know the name and the proof of the following theorem:
Let $\triangle ABC$ and $\triangle A'B'C'$ be two similar triangles,such that $P\in AA'\cap BB'\cap CC'$ .Prove that $P$ is homothety center.

if we add one condition that P is an interior point of $\triangle ABC$ , I guess the claim must be true. if anyone would post a proof for this? thanks ~

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8857 posts

#10 h Aug 28, 2022, 7:24 PM

Y by

Label the angles $\angle CDE = \angle 1$ , $\angle FDB = \angle 2$ , and so on in a clockwise order.

Assume for the sake of contradiction that $\angle 1 \neq \angle B$ , so without loss of generality let $\angle 1 > \angle B$ . Check that
$\begin{align*} \angle 1 &> \angle 4 \\ \angle 3 &> \angle 6 \\ \angle 5 &> \angle 2 \end{align*}$ by the given similarity. Then as $\angle 1 + \angle 4 = 2 \angle B < \pi$ , we have $\sin \angle 1 > \sin \angle 4$ , and similarly $\sin \angle 3 > \sin \angle 6$ and $\sin \angle 5 > \sin \angle 2$ . Thus, $\frac{CE \cdot BD \cdot AF}{CD \cdot BF \cdot AE} = \frac{\sin \angle 1 \cdot \sin \angle 3 \cdot \sin \angle 5}{\sin \angle 2 \cdot \sin \angle 4 \cdot \sin \angle 6} > 1,$ which contradicts Ceva.

Thus $\angle 1 = \angle B$ , so $\overline{DE} \parallel \overline{AB}$ . Ceva now implies $F$ is the midpoint of $\overline{AB}$ , so it follows that $P$ must be the centroid.

This post has been edited 2 times. Last edited by HamstPan38825, Aug 28, 2022, 7:25 PM

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