China TST 2000 circumcircle of triangle ADE

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China TST 2000 circumcircle of triangle ADE

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Source: China TST 2000, problem 1

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3647 posts

orl

#1 h May 22, 2005, 6:57 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle such that $AB = AC$ . Let $D,E$ be points on $AB,AC$ respectively such that $DE = AC$ . Let $DE$ meet the circumcircle of triangle $ABC$ at point $T$ . Let $P$ be a point on $AT$ . Prove that $PD + PE = AT$ if and only if $P$ lies on the circumcircle of triangle $ADE$ .

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1160 posts

#2 h May 26, 2005, 4:44 PM • 2 Y

Y by Adventure10, Mango247

I think $T$ must be between $E$ and $D$ , here I found a nice solution, but if it is not, then I am getting a contradiciton!

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#3 h Nov 16, 2005, 3:09 AM • 1 Y

Y by Adventure10

Actually, I think this point T is not well defined, as I was trying to draw the figure here...

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#4 h Nov 16, 2005, 4:19 AM • 2 Y

Y by Adventure10, Mango247

Pascual2005 wrote:

I think $T$ must be between $E$ and $D$ , here I found a nice solution, but if it is not, then I am getting a contradiciton!

Actually, I found half of it to be true when $T$ is outside the segment $DE$ , but in my drawing $D$ lies on the segment $AB$ , while $E$ liues on the ray $(CA$ , with $A$ between $E$ and $C$ . I think it might depend on this, but I'm too tired to analyze the cases. I'll just go ahead and write what my drawing is showing me

.

Suppose first that $P$ is the second intersection of $AT$ with the circle $(ADE)$ , and let $Q$ be the midpoint of the arc $DE$ of $(ADE)$ which does not contain $A$ . By Ptolemy's Theorem applied to the quadrilateral $PDQE$ , we get $PD\cdot QE+PE\cdot QD=QP\cdot DE$ , and, since $QD=QE$ and $DE=AB=AC$ , we get $(PD+PE)\cdot QD=QP\cdot AB\ (*)$ . However, the triangles $ABC,QDE$ are similar, and $\angle DQP=\angle BAT$ , while $P,T$ lie on the circumcircles of $QDE,ABC$ respectively, so the figures $QDEP,ABCT$ are similar. This gives $AT\cdot QD=QP\cdot AB$ . If we compare this with $(*)$ we get exactly what we want: $PD+PE=AT$ .

The converse is not true though, I believe. In the general case, there will be a second point $P'$ on $AT$ satisfying $P'D+P'E=AT$ . This can be seen as follows: the locus of the points $X$ with $XD+XE=AT$ is an ellipse with foci $D,E$ , passing through $P$ , the second intersection point between $AT$ and $(ADE)$ , as was shown above. In order to prove that $P$ is the only intersection point between this ellipse and $AT$ , we would have to show that $AT$ is tangent to the ellipse in $P$ or, in other words, that it is the external angle bisector of $\angle PDE$ . However, it's clear from the construction above that $PQ$ is the angle bisector of $\angle PDE$ , so the converse of what we have shown in the paragraph above holds iff $AT\perp QP$ , which is equivalent to $DE\perp BC$ , for example.

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jred

#5 h Feb 13, 2017, 3:05 PM • 2 Y

Y by Adventure10, Mango247

According to the official statement, $T$ is the intersection of segment $DE$ and $(ABC)$ , therefore $T$ is between $D$ and $E$ .

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#6 h Sep 4, 2019, 2:05 PM • 1 Y

Y by Adventure10

Does this work :help:

:help:

China TST 2000 P1 wrote:

Let $ABC$ be a triangle such that $AB = AC$ . Let $D,E$ be points on $AB,AC$ respectively such that $DE = AC$ . Let $DE$ meet the circumcircle of triangle $ABC$ at point $T$ . Let $P$ be a point on $AT$ . Prove that $PD + PE = AT$ if and only if $P$ lies on the circumcircle of triangle $ADE$ .

Solution: Suppose $P \equiv AT \cap \odot (ADE)$ . Let $F$ be a point on $PD$ such $PF=PE$ $\implies$ $\Delta FDE \cong \Delta TAC$ which follows due to angle chasing $\angle CTA=\angle ABC=\angle DFE$ ( $\Delta ABC \sim \Delta PFE$ )

Now, for the Converse, Suppose $PD+PE=AT$ . Let $P' \equiv AT \cap \odot (ADE)$ . A similar procedure shows $P'D+P'E$ $=$ $PD+PE$ $\implies$ $P' \equiv P$ $\qquad \blacksquare$

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This post has been edited 1 time. Last edited by AlastorMoody, Sep 4, 2019, 2:12 PM

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denery

#7 h Dec 18, 2020, 6:10 AM

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i have a nicer solution by length chasing
BUT failed to contradict many cases

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