Locus

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perfect_radio

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perfect_radio

#1 h Mar 8, 2005, 8:33 PM • 2 Y

Y by Adventure10, Mango247

Let $H$ be the orthocenter of the acute triangle $ABC$ . Let $BB'$ and $CC'$ be altitudes of the triangle ( $B^{\prime} \in AC$ , $C^{\prime} \in AB$ ). A variable line $\ell$ passing through $H$ intersects the segments $[BC']$ and $[CB']$ in $M$ and $N$ . The perpendicular lines of $\ell$ from $M$ and $N$ intersect $BB'$ and $CC'$ in $P$ and $Q$ . Determine the locus of the midpoint of the segment $[ PQ]$ .

Gheorghe Szolosy

This post has been edited 9 times. Last edited by perfect_radio, Mar 16, 2006, 11:09 PM

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mecrazywong

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mecrazywong

#2 h Mar 10, 2005, 6:53 PM • 2 Y

Y by Adventure10, Mango247

Note that $\{ M,P,N,B'\}$ , $\{ A,C'N,Q\}$ are concyclic, so $C'H\cdot CH=MH\cdot NH=B'H\cdot BH$ , Thus B',C',P,Q are also concyclic. $\angle HBC=\angle HC'B'=\angle HPQ$ , which implies PQ//BC. Let D be the midpoint of BC. Then the midpoint of PQ must lie on HD.
However, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\frac{2\cos A}{\cos A+1}\le\frac{HP}{HB}\le1$ . The answer is trivial then.

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darij grinberg

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#3 h Mar 10, 2005, 9:21 PM • 2 Y

Y by Adventure10, Mango247

mecrazywong wrote:

Note that $\{ M,P,N,B'\}$ , $\{ A,C'N,Q\}$ are concyclic, so $C'H\cdot CH=MH\cdot NH=B'H\cdot BH$ , Thus B',C',P,Q are also concyclic.

I think you wanted to say:

Note that the points M, P, N and B' are concyclic, and that the points N, Q, M and C' are concyclic, so that $PH\cdot B^{\prime}H=MH\cdot NH=QH\cdot C^{\prime}H$ . Thus, the points B', C', P, Q are also concyclic.

All the rest of your solution is correct and very nice.

darij

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mecrazywong

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mecrazywong

#4 h Mar 11, 2005, 9:00 AM • 2 Y

Y by Adventure10, Mango247

Darij, thanks for reminding me

and sorry for the typos

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perfect_radio

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perfect_radio

#5 h Dec 5, 2005, 9:01 PM • 2 Y

Y by Adventure10, Mango247

mecrazywong wrote:

However, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\frac{2\cos A}{\cos A+1}\le\frac{HP}{HB}\le1$ . The answer is trivial then.

I can't understand this part of the solution ....

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Virgil Nicula

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Virgil Nicula

#6 h Dec 12, 2005, 10:13 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

This problem is a "painting" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\parallel BC$ . The its prove is identically with the Darij's solution. It is a pity for the author of the initial problem!

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perfect_radio

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perfect_radio

#7 h Dec 12, 2005, 10:41 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula, do you know what mecrazywong meant by that trig at the end of his solution?

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mecrazywong

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mecrazywong

#8 h Dec 26, 2005, 8:00 PM • 2 Y

Y by Adventure10, Mango247

We are finding the locus of the midpoint of PQ. The first part of my solution only proves that the midpoint of PQ lies on HD. Since not every points on HD does belong to the family of the loci, we must eliminate them and the trigo inequality at the end of my solution serves for this purpose.

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perfect_radio

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perfect_radio

#9 h Dec 26, 2005, 9:17 PM • 2 Y

Y by Adventure10, Mango247

mecrazywong, i also deduced that. but how did you reach that trigo inequality? that was what i was asking

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perfect_radio

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perfect_radio

#10 h Feb 14, 2006, 6:10 PM • 2 Y

Y by Adventure10, Mango247

perfect_radio wrote:

mecrazywong, i also deduced that. but how did you reach that trigo inequality? that was what i was asking

?

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Virgil Nicula

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Virgil Nicula

#11 h Feb 14, 2006, 7:06 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

This problem is a "painting" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\parallel BC$ . The its prove is identically with the Darij's solution. It is a pity for the author $\boxed {S.\ J.\ Weyl}$ of the initial problem!

This problem (R.M.O. $\boxed {2005}\ ,\ 9^{th}$ -grade) appears in my book (only with $PQ\parallel BC$ , without geometrical locus) " Geometrie plana. Culegere de probleme, Ed. GIL, $\boxed {2002}$ (the page 51 with the number 5.17). Am I right ? $\boxed {\ 2005\ -\ 2002\ =\ 3\ >\ 0\ }$ Question: Who is the author of this problem from R.M.O., 2005 ?

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perfect_radio

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perfect_radio

#12 h Feb 26, 2006, 4:45 PM • 2 Y

Y by Adventure10, Mango247

I don't know who is the author of this problem. My teacher gave me a year ago a "Gazeta Matematica", which contained the problems from RMO 2004 and their authors. But he took it back shortly after.

Do you know how mecrazywong reached that trig inequality? I can't seem to obtain anything similar. (The other part with $PQ \| BC$ I understood)

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armpist

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armpist

#13 h Feb 26, 2006, 9:05 PM • 2 Y

Y by Adventure10, Mango247

The limiting position of midpoint of PQ happens when

it is colliner with the radical axis of the Droz-Farny cirles.

T.Y.

M.T.

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