nasty geometry problem

[siyan]

PR and PQ are chords of a circle with center O. OT is perpendicular to PQ and OS is perpendicular to PR. If OT = OS, prove that T, S, R, and Q are concyclic.

That had me going for hours....any help appreciated.

-C

[krtxmrtz]

How do you define the lines OT and OS, i.e. are the points T and S located on the chords or where?

[DiGiTaIErRoR]

Got a picture?

[siyan]

Originally posted by krtxmrtz
How do you define the lines OT and OS, i.e. are the points T and S located on the chords or where?

it doesn't say....which is exactly why i think it can't be done...

-C

[siyan]

Originally posted by DiGiTaIErRoR
Got a picture?

no pic. the textbook gave me exactly what i gave you

[sql_lall]

OK, a few things:
1) I don't think it matters where T and S are, so long as T lies on the line OQ (line, not line segment), and S lies on the line RO (again, line, not line segment), and that OT = OS

However, i don't think it can be done because of this case:

1) Construct a circle centre O
2) choose a point P outside O, and join the 2 chords PR and PS
3) Join RO and SO.
4) Place T, between the circumference of the circle the 'other' side of Q, and O (i.e. TOQ is the smaller than diameter, but bigger than the radius.)
5) Place S on the line segment RO, so that OS = OT.
6) RQST are definitely not cyclic, as S is inside RQT

[NotLKH]

Hmmm, I'm starting to see something.

OK, Is the Following Image a possible illustration for the problem?



And, since OT and OS are perpindicular to PQ and PR respectively,
Then OT and OS Bisect PQ and PR?

Am I right so Far?

[Guv]

A radius perpendicular to a chord is a perpendicular bisector. Id est: The radius crosses the chord at the midpoint.

If S & T are points on the chords and OT = OS, then the chords are equal.

If S & T are not on the chords, then I am not sure how to proceed.

What is concyclic?

[bugzpodder]

if a quadrilateral is said to be concyclic, it is possible to draw a circle through all 4 vertices of the quadrilateral.

[NotLKH]

Originally posted by Guv
If S & T are points on the chords and OT = OS, then the chords are equal.

True, thats how I was initially interpreting it.
But, assume T and S are not on the chords.

ConCyclic means that given 4 points, if All 4 satisfies the equation of a single circle, then they are concyclic.

ie... If you can find a single point equidistant from all 4, than they are concyclic, and that point is the center of the circle that they lie upon.

So, I beleive if we draw perpindicular bisectors of RS, St, and TQ, and prove that those three bisectors Meet at a single point, then we've proven that R, S, T, and Q are concyclic.

-Lou

[bugzpodder]

So, I beleive if we draw perpindicular bisectors of RS, St, and TQ, and prove that those three bisectors Meet at a single point, then we've proven that R, S, T, and Q are concyclic.

there are distinctive properties of concylic quadrilaterals (such as opposite angles add up to 180), but i am not sure if the above is a property of a concyclic quadrilateral.

[NotLKH]

Well, if Concyclic points are {4 or more } points that lie on a single circle, then
if you view them as end points of chords of the same circle, then
obviously if you draw perpindicular bisecters of them then they'd
all pass thru the center of the circle, hence they'd meet at the
center of the circle.

[bugzpodder]

yes i agree.

[krtxmrtz]

Wait a minute!!!

Are PR and PQ required to have the same length??? I have been fooling around with this for quite a while assuming they aren't. Then I have tried an example geometrical construction on paper with PQ<>PR and the 4 points turn out to be definitely NOT concyclic. That's why I ask...

Originally posted by siyan
PR and PQ are chords of a circle with center O. OT is perpendicular to PQ and OS is perpendicular to PR. If OT = OS, prove that T, S, R, and Q are concyclic.

That had me going for hours....any help appreciated.

-C

[krtxmrtz]

OK, I'm sure the chords must be of the same length for I have finally worked out the proof assuming this.

I'll be posting it as soon as have some time, probably in a few hours or tomorrow at the latest.

[krtxmrtz]

Originally posted by krtxmrtz
OK, I'm sure the chords must be of the same length for I have finally worked out the proof assuming this.

I'll be posting it as soon as have some time, probably in a few hours or tomorrow at the latest.

Here it is.

There might be a more straighforward way to prove it using only geometry but I haven't been able to find it so, I have finally taken to brute force i.e. algebra.

Referring to the attached drawing, let's assume there really exists a circumference going through all 4 points P,Q,R and S. Call R its radius and O its origin, with coordinates (a,b). The way I've done it is, using the known data (r, w and t), I write down these 4 equations:

(1) Distance OP = R
(2) Distance OQ = R
(3) Distance OS = R
(4) Distance OT = R

Now, that's 4 equations for 3 unknowns: a, b and R. We know that there is always one circumference going through any 3 points, which means if we take a subset of 3 of the above equations we could solve them to find a unique solution for a, b and R. Then we can introduce these 3 values in the fourth equation and verify whether or not the equality holds, i.e. whether or not the fourth point lies on the circumference.

Coordinates of the points:

P: (r,0)
Q: (rcos(2w),rsin(2w))
S: (tcos(w/2),tsin(w/2))
T: (tcos(3w/2),tsin(3w/2))

where I have assumed an arbitrary length t for the segments OS and OT.

The above 4 equations are then:

(1) (r-a)²+b²=R²
(2) (rcos(2w)-a)²+(rsin(2w)-b)²=R²
(3) (tcos(w/2)-a)²+(tsin(w/2)-b)²=R²
(4) (tcos(3w/2)-a)²+(tsin(3w/2)-b)²=R²

where the squared distances rather than the distances themselves are written.

From (1):

(5) r²+a²+b²=R²+2ra

Squaring the parentheses in (2):

r²cos²(2w)+a²-2racos(2w)+r²sin²(2w)+b²-2rbsin(2w)=R²

which becomes,

r²+a²+b²-2r(acos(2w)+bsin(2w))=R²

Substituting (5) into the latter equation:

2ra-2racos(2w)-2rbsin(2w)=0

from which we arrive at:

(6) b=a(1-cos(2w))/sin(2w)

Carrying out the squares in (3) and (4) we have:

(7) s²+a²+b²-2t(acos(w/2)+bsin(w/2))=R²
(8) s²+a²+b²-2s(acos(3w/2)+bsin(3w/2))=R²

Substituting (7) into (8):

(9) acos(w/2)+bsin(w/2)-acos(3w/2)-bsin(3w/2)=0

I now substitute (6) into (9):

cos(w/2)+sin(w/2)(1-cos(w/2))/sin(2w)-cos(3w/2)-sin(3w/2)(1-cos(2w))/sin(2w)=0

and removing the denominators, we are left with:

(10) sin(2w)cos(w/2)+sin(w/2)(1-cos(2w))-cos(3w/2)sin(2w)-sin(3w/2)(1-cos2w))=0

So, if this can be shown to be true, than we're done.

Because I don't like to work with w/2, I now define a new variable: v=w/2. Then:

sin(4v)cos(v)+(1-cos(4v))sin(v)-sin(4v)cos(3v)-(1-cos(4v))sin(3v)=0

sin(4v)cos(v)+sin(v)-cos(4v)sin(v)-sin(4v)cos(3v)-sin(3v)+cos(4v)sin(3v)=0

If I now use the well known trigonometric formula:

sin(x-y)=sin(x)cos(y)-sin(y)cos(x), I finally have:

sin(4v)cos(v)-sin(v)cos(4v)=sin(4v-v)=sin(3v)
and
sin(3v)cos(4v)-sin(4v)cos(3v)=sin(3v-4v)=sin(-v)=-sin(v)

so at the end all terms cancel out and the equality is true.

QED

[NotLKH]

Originally posted by krtxmrtz
Here it is.

.....
Referring to the attached drawing, let's assume there really exists a circumference going through all 4 points P,Q,R and S. Call R its radius and O its origin, with coordinates (a,b). The way I've done it is, using the known data (r, w and t), I write down these 4 equations:

(1) Distance OP = R
(2) Distance OQ = R
(3) Distance OS = R
(4) Distance OT = R

Now, that's 4 equations for 3 unknowns: a, b and R. We know that there is always one circumference going through any 3 points, which means if we take a subset of 3 of the above equations we could solve them to find a unique solution for a, b and R. Then we can introduce these 3 values in the fourth equation and verify whether or not the equality holds, i.e. whether or not the fourth point lies on the circumference.
....

I've got a couple of questions at this point.
First of all, Why do you use R to represent the radious for the points P,Q,R, and S? R is already defined as a Point, so it gets very confusing.

Num2: O is already the center of the original Circle which ahs the points R, Q, and P residing on its circumference. Why do you reuse O to define the center of the 2nd hypothetical circle which we are trying to prove exists with points Q, R, S, and T riding on it's circumference. Again, confusing.

Num3: When you say:

(1) Distance OP = R
(2) Distance OQ = R
(3) Distance OS = R
(4) Distance OT = R

And just for clarity, let me substitute O₂ for the center of the second hypothetical circle we are trying to prove, and R₂ for the radius of the second hypothetical circle we are trying to prove, shouldn't it actually be said that:

(1) Distance O₂Q = R₂
(2) Distance O₂R = R₂
(3) Distance O₂S = R₂
(4) Distance O₂T = R₂

It is never said that P resides on the same circle with Q,R,S, and T.

And, let me attach a second Image, to furthur elucidate what I think we are trying to prove, and please tell me if you agree or not.



I believe we are trying to prove the existence of the blue circle {Center O₂ and radius Rad₂}, with points Q,R,S, and T riding on its circumference, satisfying the original conditions.

[krtxmrtz]

Originally posted by NotLKH
I've got a couple of questions at this point.
First of all, Why do you use R to represent the radious for the points P,Q,R, and S? R is already defined as a Point, so it gets very confusing.

You're right, sorry. I should have used R₂ or rename point R to point ... G?

Num2: O is already the center of the original Circle which ahs the points R, Q, and P residing on its circumference. Why do you reuse O to define the center of the 2nd hypothetical circle which we are trying to prove exists with points Q, R, S, and T riding on it's circumference. Again, confusing.

Again, I agree. I was working on the second circle and just forgot about O being already used. So, let's use O₂ as you suggest.

Num3: When you say:

And just for clarity, let me substitute O₂ for the center of the second hypothetical circle we are trying to prove, and R₂ for the radius of the second hypothetical circle we are trying to prove, shouldn't it actually be said that:

It is never said that P resides on the same circle with Q,R,S, and T.

Here, to add yet to the confusion, I wrote P when I really meant R. Sorry again.

And, let me attach a second Image, to furthur elucidate what I think we are trying to prove, and please tell me if you agree or not.



I believe we are trying to prove the existence of the blue circle {Center O₂ and radius Rad₂}, with points Q,R,S, and T riding on its circumference, satisfying the original conditions.

Thanks for pointing out my mistakes, one should really rush only to go home from work.

[A$$Bandit]

"One" should? You know it's forbidden by international law for anyone who isn't British to use the word "one" as a pronoun...

[Guv]

Without serious analysis, the following statements seem obviously valid.

A radial line from the center of the circle and perpendicular to a chord intersects the chord at the midpoint.

As stated, the two chords (PQ & PR) are not necessarily the same length, but this possibility is not excluded. Krtxmrtz: You seem to think they are equal, but I do not see that stated in the origianl post.

As stated, OT = OS, but the length of them is otherwise arbitrary.

If the problem is to find a length such that the conclusion is true, then that length is the radius of the original circle.

From the above is seems that the problem implies that there is an unbounded number of points S & T which fit the conditions and are concyclic.

Perhaps it might be worthwhile to pick a specific convenient value for the length OS=OT and try to prove or disprove the conclusion for that particular case.

A disproof for a particular value of OS=OT would indicate that the conclusion as stated is not true. A disproof for a particular OS=OT and equal chords of a specific length would also invalidate the conclusion as stated. Proofs for specific values would reinforce the conclusion and might indicate how to construct a general proof.

Perhaps somebody with accurate drafting tools can verify or refute the conclusion, noting that any arbitrary length OS=OT fits the conditions, even values less than the radious of the origianl circle..

[krtxmrtz]

Originally posted by Guv
Without serious analysis, the following statements seem obviously valid.

A radial line from the center of the circle and perpendicular to a chord intersects the chord at the midpoint.

That IS true, indeed (can't remember though if it's a more or less obvious conclusion from some basic geometry principles or if it needs some sophisticated derivation...)

As stated, the two chords (PQ & PR) are not necessarily the same length, but this possibility is not excluded. Krtxmrtz: You seem to think they are equal, but I do not see that stated in the origianl post.

I think it is not stated because Siyan may have forgotten about doing it. I am convinced that the statement is only true if the chords have the same length. As a matter of fact, I'll try to prove in another (future) post that if those chords do NOT have the same length, then in no case can the points be concyclic.

[krtxmrtz]

Btw, I have finally come out with the easy elegant way!

The geometrical proof was so easy and self-evident that I am ashamed of myself for having overlooked it for about 4 days.

Looking at the 4 figures:

1. Without loss of generality, take point P on the x axis, so that points Q and R are symmetrical with respect to it.

2. That means the center O' of the cincumference that's supposed to contain Q, R, S and T lies on the x axis, for it has to be on the straight line perpendicular to QR.

3. Likewise, O' will lie on the straight line perpendicular to QS and will be located at its intersection with the x axis.

4. Finally, the same construction can be made for points T and R. By symmetry, the straight line perpendicular to TR will go through O' thus proving that T is also concyclic.

QED (2)

You know, Siyan, it wasn't such a nasty problem after all...!