Extend DP until it cuts the circumcircle again at N

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Extend DP until it cuts the circumcircle again at N

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Source: China TST 1993, problem 6

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3647 posts

orl

#1 h Jun 27, 2005, 10:35 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ be a triangle and its bisector at $A$ cuts its circumcircle at $D.$ Let $I$ be the incenter of triangle $ABC,$ $M$ be the midpoint of $BC,$ $P$ is the symmetric to $I$ with respect to $M$ (Assuming $P$ is in the circumcircle). Extend $DP$ until it cuts the circumcircle again at $N.$ Prove that among segments $AN, BN, CN$ , there is a segment that is the sum of the other two.

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#2 h Jun 27, 2005, 2:29 PM • 2 Y

Y by Adventure10, Mango247

http://www.mathlinks.ro/Forum/viewtopic.php?t=15773 .

darij

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3647 posts

orl

#3 h Jun 27, 2005, 3:43 PM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem.

Congrats for making a big step for being on the Chinese team.

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3498 posts

#4 h Jun 28, 2005, 10:41 AM • 2 Y

Y by Adventure10, Mango247

orl wrote:

shobber wrote:

Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem.

Congrats for making a big step for being on the Chinese team.

I got this information from a book. It provided a elementary trig proof, which is also hard to come up with.
The book says we can turn this problem into proving:
$\tan{\angle{ADN}}=\frac{\sin{C}-\sin{B}}{1-\cos{B}-\cos{C}}$
Most guys managed to get here, but failed to go deeper.

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#5 h Jun 29, 2005, 8:08 AM • 2 Y

Y by Adventure10, Mango247

A very hard and nice problem! I have nearly spent 2 hours on it.

WLOG, assume $b\ge a\ge c$ . We will start with 2 lemmas:
Lemma 1
$\angle ACN=\angle IOD$ .
Proof:
Let D' be the reflection of D in BC, D'' the antipode of D.
By similar triangle, $BD^2=DM\cdot DD''\Rightarrow DI^2=2DM\cdot DO=DD'\cdot DO$ . Thus, $\angle IOD=\angle D'ID=\angle PID+\angle IPD=180^\circ-\angle IDP=\angle ACN$ .

Lemma 2
Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$
Proof: problem posted before by me

Now we apply a homothety to $\triangle AXY$ centered at A with the ratio $\frac{AC}{AY}$ . Suppose the image is ACZ. Then consider the spiral smilarity of $\triangle ACZ$ centered at C through an angle $\angle ZCB$ (anticlockwisely) and ratio $\frac{BC}{AC}$ . Suppose the image is $\triangle CBN'$ . Clearly $N'\in(O)$ . We have $\frac{BN'}{CN'}=\frac{a-c}{b-a}\Rightarrow bBN'+cCN'=a(BN'+CN')$ . But the Ptolemy Theorem yields $bBN'+cCN'=aAN'$ . Therefore BN'+CN'=AN'.

Note that $90^\circ=\angle(XY,IO)=\angle(ZC,IO)=\angle(BC,IO)+\angle BCZ=90^\circ-\angle IOD+\angle ACN'$ . Thus, $\angle IOD=\angle ACN'$ . By Lemma 1, N=N', so we also have AN=BN+CN.
For the case $b\ge c\ge a,a\ge b\ge c$ , etc. The proof is essentially the same so I don't want to repeat, which is extremely tedious and boring.

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#6 h Feb 2, 2014, 4:10 PM • 2 Y

Y by Adventure10, Mango247

mecrazywong wrote:

Lemma 2
Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$
Proof: problem posted before by me

Do you have a link or idea for proof this lemma ?

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kazi

#7 h Feb 8, 2014, 2:31 PM • 2 Y

Y by Adventure10, Mango247

I have a really nice solution . I made it yesterday
as all know we must prove $\sin AP=\sin BP + \sin CP$
we know that $\frac{XM}{XI}=\frac{1}{2}=\frac{TM}{TI}(=\sin \frac{A}{2}).\frac{\sin AP -(\frac{B-C}{2})}{\sin AP}\Leftrightarrow \sin AP = 2.\sin \frac{A}{2}.\sin AP -(\frac{B-C}{2})=\sin PB + \sin PC$

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jred

#8 h Feb 9, 2017, 3:33 AM • 2 Y

Y by Adventure10, Mango247

In early 90's, this problem seemed very hard, but it's not that hard nowaday.
I saw a very smart and simple solution, maybe the official solution, on a Chinese journal by using the area means.

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jayme

#9 h Feb 9, 2017, 8:36 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
I have also this reference

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323883

Sincerely
Jean-Louis

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DottedCaculator

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DottedCaculator

#10 h Jan 29, 2024, 6:52 PM

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Let $a=x^2$ , $b=y^2$ , $c=z^2$ , $d=-yz$ , $i=-xy-yz-zx$ , $m=\frac{y^2+z^2}2$ , $p=y^2+z^2+xy+yz+zx$ . Then,

$\begin{align*} n&=\frac{d-p}{d\overline p-1}\\ &=\frac{-(y+z)(x+y+z)}{-yz\left(\frac1{y^2}+\frac1{z^2}+\frac1{xy}+\frac1{yz}+\frac1{zx}\right)-1}\\ &=\frac{-xyz(y+z)(x+y+z)}{(y+z)(xy+yz+zx)}\\ &=\frac{xyz(x+y+z)}{xy+yz+zx}. \end{align*}$
Therefore,
$n-a=\frac{xyz(x+y+z)-x^2(xy+yz+zx)}{xy+yz+zx}=\frac{x(yz-x^2)(y+z)}{xy+yz+zx},$ so $AN=\frac{|yz-x^2||y+z|}{|xy+yz+zx|}.$ Therefore, $|yz-x^2|^2|y+z|^2=-\frac{(yz-x^2)^2(y+z)^2}{x^2y^2z^2}$ , so $ANxyz|xy+yz+zx|=\pm i(yz-x^2)(y+z)$ . Therefore, there exists a sign $\varepsilon_A$ such that $ixyz|xy+yz+zx|\varepsilon_AAN=y^2z+yz^2-x^2y-x^2z$ , so taking the cyclic sum implies $\varepsilon_AAN+\varepsilon_BBN+\varepsilon_CCN$ , so one of the three segments is the sum of the other two.

This post has been edited 1 time. Last edited by DottedCaculator, Jan 29, 2024, 6:54 PM

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