MHB Welcome to the Math Challenge Board

AI Thread Summary
A new member introduces themselves to the Math Challenge Board, expressing their intention to post challenging math problems for others to solve. The initial problem involves proving that segments PR and QS are equal under specific conditions related to points and lines. Participants engage in clarifying the problem, discussing the implications of perpendicular bisectors and the intersection of lines with circles. After some back-and-forth, a solution is proposed involving geometric relationships and angles in a semicircle. The discussion highlights collaborative problem-solving and the sharing of mathematical insights.
Albert1
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HI : good to see you everybody ,I am a new comer on this board
Being a math teacher ,I always trained my students with various difficult problems
from now on I am going to post a sequential challenging questions for people to share
most of them I know the answer and the solution of it , maybe your solution is quicker and smarter (Yes)
please try it many thanks

here comes the problem :

28l6yj9.jpg

 
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Re: Prove PR=QS

Hi Albert, :)

The second sentence in your question isn't clear to me.

a line passing through point B and intersects two points \(P,\,Q\) with \(O\) and intersects \(R,\, S\) with \(O_1\) and \(O_2\), respectively prove: \(\overline{PR} = \overline{QS}\)

There are infinitely many lines passing though \(B\) and not all of them will have \(PR = QS\). Can you please elaborate a bit more on what you mean by this sentence.

Kind Regards,
Sudharaka.
 
Re: Prove PR=QS

Can you plot a line passing through point B ,
satisfying the restriction and show that PR and QS are not equal ?
 
Re: Prove PR=QS

Albert said:
Can you plot a line passing through point B ,
satisfying the restriction and show that PR and QS are not equal ?

I might be missing something here but what is meant by "the restriction"? Do you mean that \(O\) lies on the perpendicular bisector of \(PQ\) ?
 
Re: Prove PR=QS

Sudharaka said:
I might be missing something here but what is meant by "the restriction"? Do you mean that \(O\) lies on the perpendicular bisector of \(PQ\) ?
yes, that must be
 
Re: Prove PR=QS

Albert said:
yes, that must be

But in that case there are infinitely many lines as I have claimed above, since for every chord of a circle the perpendicular bisector passes through the center of the circle. Refer >>this<<.
 
Re: Prove PR=QS

the line passing through point B must also intersect with those of two small circles :
O1,and O2, at point R and S respectively as mentioned before
 
Re: Prove PR=QS

Albert said:
the line passing through point B must also intersect with those of two small circles :
O1,and O2, at point R and S respectively as mentioned before

I think we are misunderstanding each other. To clarify further I have created this diagram using http://www.geogebra.org/cms/ and drawn two possible lines for \(PQ\).

w1vaep.jpg
 
Re: Prove PR=QS

where is your points R1,S1 and R2,S2
we want to prove P1R1=Q1S1 , and P2R2=Q2S2
 
  • #10
Re: Prove PR=QS

I understand the question now. It seems that for all lines drawn through \(B\), \(PR=QS\). Sorry for the confusion. :)

I haven't found the proof yet, but I am trying... :confused:
 
  • #11
Re: Prove PR=QS

geom.jpg
In the diagram, all the angles marked as right angles are angles in a semicircle (and therefore are indeed right angles). The two angles marked $\alpha$ are equal, as are the two angles marked $\beta$ (angles in the same segment). From the triangle $APC$, $AP = AC\cos\alpha$. From the triangle $ARP$, $PR = AP\cos\beta$. So $PR = AC\cos\alpha\cos\beta.$

In the same way, $CQ = AC\cos\beta$ (from triangle $AQC$) and $SQ = CQ\cos\alpha$ (from triangle $CSQ$), and so $SQ = AC\cos\alpha\cos\beta = PR.$
 
  • #12
Re: Prove PR=QS

Opalg:very good solution (Yes)
 
  • #13

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