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.

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$

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.

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

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

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

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...

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 ~

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.