tiring... [two equal chords AB and CD; < ALC = 2 < MON]

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tiring... [two equal chords AB and CD; < ALC = 2 < MON]

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Source: Belarus 1999

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ARTI

#1 h Mar 1, 2005, 9:50 PM • 2 Y

Y by Adventure10, Mango247

Let O be the center of circle W. Two equal chords AB and CD of W intersect at L such that AL>LB and DL>LC. Let M and N be points on AL and DL respectively such that (ALC)=2*(MON). Prove that the chord of W passing through M and N is equal to AB and CD.

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Armo

#2 h Mar 2, 2005, 9:20 PM • 2 Y

Y by Adventure10, Mango247

$(XYZ)=$ (area of $XYZ$ )?? or what?

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ARTI

#3 h Mar 3, 2005, 8:19 PM • 2 Y

Y by Adventure10, Mango247

Armo wrote:

$(XYZ)=$ (area of $XYZ$ )?? or what?

ooops...I'm sorry! (ALC),(MON)-angles

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Armo

#4 h Mar 3, 2005, 8:39 PM • 2 Y

Y by Adventure10, Mango247

Heh... That's why it was so difficult (I thought them areas).

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#5 h Mar 3, 2005, 8:57 PM • 2 Y

Y by Adventure10, Mango247

Armo wrote:

Heh... That's why it was so difficult (I thought them areas).

OK!

...And now what do you think about it

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#6 h Mar 4, 2005, 1:19 PM • 2 Y

Y by Adventure10, Mango247

Problem. Let W be a circle, and AB and CD two equal chords of this circle. Let these two chords AB and CD intersect at a point L such that AL > LB and DL > LC. Finally, let M and N be two points on the segments AL and DL, respectively, such that < ALC = 2 < MON, where O is the center of the circle W. Prove that the chord of the circle W passing through the points M and N is equal to the chords AB and CD.

Solution. Let's start with a lemma:

Lemma 1. If ABC is a triangle, and P is a point in its plane such that:

- this point P lies on the angle bisector of the angle CAB;
- we have $\measuredangle BPC=\frac{180^{\circ}-\measuredangle CAB}{2}$ ;
- the points A and P lie in different halfplanes with respect to the line BC,

then the point P must be the A-excenter of the triangle ABC (i. e., the excenter opposite to the vertex A).

Proof of Lemma 1

Now let's solve the problem itself:

Let the line MN intersect the circle W at two points Q and R. Then, the chord QR of the circle W is the chord passing through the points M and N. And we have to prove that QR = AB = CD.

Since < ALC = 2 < MON, but < ALC = 180° - < NLM, we have 2 < MON = 180° - < NLM, so that $\measuredangle MON=\frac{180^{\circ}-\measuredangle NLM}{2}$ . Moreover, it is clear that the points L and O lie in different halfplanes with respect to the line MN.

Two chords of a circle are equal if and only if the center of the circle has equal distances to these two chords. Hence, as AB and CD are two equal chords in the circle W, the center O of the circle W has equal distances from these two chords AB and CD; in other words, the point O has equal distances from the lines ML and NL, and consequently, this point O lies on the angle bisector of the angle NLM.

So let's combine our results:

- The point O lies on the angle bisector of the angle NLM.
- We have $\measuredangle MON=\frac{180^{\circ}-\measuredangle NLM}{2}$ .
- The points L and O lie in different halfplanes with respect to the line MN.

Hence, by Lemma 1, applied to the triangle LMN, it follows that the point O is the L-excenter of the triangle LMN. Thus, the point O has equal distances from the sidelines MN, NL and LM of this triangle. In other words, the point O has equal distances from the chords QR, CD and AB of the circle W. But remembering again that two chords of a circle are equal if and only if the center of the circle has equal distances to these two chords, we see that, since the point O is the center of the circle W, the equality of the distances of the point O from the chords QR, CD and AB yields the equality of these chords. In other words, we have QR = AB = CD. This solves the problem.

Darij

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#7 h Mar 4, 2005, 1:41 PM • 2 Y

Y by Adventure10, Mango247

We start from the converse of the statement: Prove that if the chord passing through M,N equals to AB and CD, then $\angle ALC=2\angle MON$ .
Let X,Y,Z be the feet of the perpendicular from O to lines AB,CD,MN respectively. Then OX=OY=OZ, which implies $\angle XOM=\angle ZOM,\angle YON=\angle ZON$ . By simple angle chasing, we can conclude that $\angle ALC=2\angle MON$ .
Now we go back to the original problem. We first choose the position of M. Then clearly the position of N is fixed, which solves the problem.

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#8 h Dec 24, 2022, 9:20 AM

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Let $O$ be the center of circle $W$ . Two equal chords $AB$ and $CD$ of $W$ intersect at $L$ such that $AL>LB$ and $DL>LC$ . Let $M$ and $N$ be points on $AL$ and $DL$ respectively such that $\angle ALC=2 \angle MON$ . Prove that the chord of $W$ passing through $M$ and $N$ is equal to $AB$ and $CD$ .

This post has been edited 3 times. Last edited by parmenides51, Jan 7, 2023, 11:55 PM

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