Small question..

Fruityloop · 2009-10-22 16:37:29

This is from a problem book...

AB and CD are perpendicular diameters of a circle and are 10 units in length.
Chord CH cuts AB at K and is 8 units long.

Let x = KB and y = CK.
The book says

Why?

bobbym · 2009-10-23 02:47:13

Hi Fruityloop;

I used analytical geometry and algebra:

Graph the circle:

The x and y axes will act as the perpendicular diameters. Label (-5,0) as C, (5,0) as D, (0,5) as A and (0,-5) as B.
.

If you draw the chord that has the equation ( 3 / 4 ) x + (15 / 4 ) you will pass through C and intersect the circle at ( 7 / 5 , 4.8). This chord will have a length of 8. Label the point where the chord intersects the y axis ( AB ) as K. And where it touches the circle as H. The chord intersects the y axis ( AB ) at (0,3.75) labelled K.

Fruityloop wrote:

Chord CH cuts AB at K and is 8 units long.

I have presumed that the 8 units represents the the length of CH (the chord) and not CK ( CK is not a chord).

You now have everything you need to get all the segments by pythagoras. After getting them:

x = 8.75
(10 - x ) = 1.25
y = 6.25
( 8 - y ) = 1.75

8.75 * 1.25 = 6.25 * 1.75

Last edited by bobbym (2009-10-23 04:48:55)

Fruityloop · 2009-10-23 11:16:15

Thank you for your time and effort Bobbym. So it seems that it's just a happy coincidence of this particular problem. I thought there was some great theorem that I wasn't aware of that allowed this to be true. Pretty funny. After going through your analysis I see how to solve this. Thank you.

Fruityloop

bobbym · 2009-10-23 11:40:30

Hi Fruityloop;

That I can't say. This is like a particular solution. There may be a general law here, I don't know. Anyway, glad to be of some help.

soroban · 2009-10-23 13:06:00

.

              * * *
       C  *           *
        o               *
       * \               *
          \y      
      *    \ K      x     *
    A o - - o - - - - - - o B
      * 10-x \            *
              \   
       *       \ 8-y     *
        *       \       *
          *      \     *
              * * o
                  H

.

Identity · 2009-10-23 15:34:25

Yep, that is a fantastic theorem to remember for other problems, thanks for the derivation soroban

Fruityloop · 2009-10-26 16:25:07

I've been looking for some proof of this and I think I've found it..
Euclid's Elements Book III proposition 21 states..
"In a circle, the angles in the same segment equal one another."
So in Soroban's diagram, if we draw lines connecting A to C, C to B, B to H, and H to A
we immediately see that

.
We consequently end up with similar triangles and the segments of each line when multiplied together end up being equal to each other.
So the expression is true for ANY four points on the circumference on a circle. It seems this has been known for only about 2300 years!

Well, live and learn.
Fruityloop

Last edited by Fruityloop (2009-10-26 16:31:40)

bobbym · 2009-10-27 03:45:54

Hi Fruityloop;

I stuck to the terms of the original question and drew the chord from C, but it is easy to see that you could draw the chord from A,B,C or D covering all possible chords.

When I started this I was convinced that it would be shorter than the euclidean geometry proof because often coordinate geometry proofs are. That is not the case here.

So it is proved for all chords in the diagram.

Last edited by bobbym (2009-10-28 03:09:51)

Sarah12 · 2009-10-29 09:43:57

Hum That Sounds. Geometry

Math Is Fun Forum

#1 2009-10-22 16:37:29

Small question..

#2 2009-10-23 02:47:13

Re: Small question..

#3 2009-10-23 11:16:15

Re: Small question..

#4 2009-10-23 11:40:30

Re: Small question..

#5 2009-10-23 13:06:00

Re: Small question..

#6 2009-10-23 15:34:25

Re: Small question..

#7 2009-10-26 16:25:07

Re: Small question..

#8 2009-10-27 03:45:54

Re: Small question..

#9 2009-10-29 09:43:57

Re: Small question..

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