1. Problem with a chord

A chord of length 18 is perpendicular to a diameter at a point A. If the circle diameter is 30, find the number of chords of integral length (other than the diameter) that contain A.

I don't know how to draw this and completely stuck if you can help and shows how to I would appreciate it. Thanks

2. Re: Problem with a chord

Originally Posted by stevecowall
A chord of length 18 is perpendicular to a diameter at a point A. If the circle diameter is 30, find the number of chords of integral length (other than the diameter) that contain A.

I don't know how to draw this and completely stuck if you can help and shows how to I would appreciate it. Thanks
1. Draw a sketch. (see attachment)

2. You are dealing with a right triangle whose height is 9.

According to the Euclidean geometry the product of the two hypotenuse segments equals the square of the height.

That means you have to solve for x:

$9^2=x(30-x)~\implies~x^2-30x+81=0$

3. You'll get two integer results which add up to 30.

3. Re: Problem with a chord

Originally Posted by stevecowall
A chord of length 18 is perpendicular to a diameter at a point A. If the circle diameter is 30, find the number of chords of integral length (other than the diameter) that contain A.

I don't know how to draw this and completely stuck if you can help and shows how to I would appreciate it. Thanks
Second attempt (only in case I misunderstood the question)

1. Draw a sketch (see attachment)

2. According to the result of my previous post the diameter is divided into 2 segments of 3 or 27 length.

3.According to the intersecting chord theorem you have to find integer values for x and y if

$x \cdot y = 3 \cdot 27 = 81~\implies~y=\frac{81}{x}$

4. There are only a few triples of numbers possible:

$(x,y,length):\left \{ \begin{array}{l}(1,81,82)\\(3,27,30)\\(9,9,18)\\(2 7,3,30)\\(81,1,82)\end{array} \right.$

5. That means there is only 1 chord which satisfies the given conditions.

4. Re: Problem with a chord

Hmm, I think the problem is a bit simpler than that.
It seems to me that the chords of integral length that contain A do not require the chord to exists of 2 integral sections with A as end point.
Nor do they have to be perpendicular to the original diameter.

5. Re: Problem with a chord

I agree. I think that as the angle between the chord and the diameter changes from 90 to 0 degrees, the chord length changes continuously from its shortest value 18 to its longest value 30.

6. Re: Problem with a chord

Originally Posted by stevecowall
A chord of length 18 is perpendicular to a diameter at a point A. If the circle diameter is 30, find the number of chords of integral length (other than the diameter) that contain A.

I don't know how to draw this and completely stuck if you can help and shows how to I would appreciate it. Thanks
Originally Posted by emakarov
I agree. I think that as the angle between the chord and the diameter changes from 90 to 0 degrees, the chord length changes continuously from its shortest value 18 to its longest value 30.
If you are right then the question can be answered by counting:

Let l denotes the length of the chord and $l \in \mathbb{N}$. Then you know:

$18\leq l < 30$

So you get 12 values for l.

7. Re: Problem with a chord

Originally Posted by earboth
If you are right then the question can be answered by counting:

Let l denotes the length of the chord and $l \in \mathbb{N}$. Then you know:

$18\leq l < 30$

So you get 12 values for l.
I made the answer 23. There are two chords of each integral length (one with positive gradient and one with negative gradient), with the exception of the chord of length 18, which is unique.

8. Re: Problem with a chord

On another forum, the chord perpendicular to a diameter is given by the OP as being 16 units in length...and it is not specified whether we discard the given chord, only to discard the diameter as a chord. The OP states the answer is 1 to 26, which I found is the answer if we discard the diameter and the given chord:

http://http://www.mymathforum.com/vi...104315#p104315

9. Re: Problem with a chord

It was my mistake of posting ... It supposes to be 18

10. Re: Problem with a chord

In that case we find 2 chords each for a length of {19,20,...,28,29} = 22 chords. We have excluded the given chord and the diameter. Your answer of 26 on the other site only makes sense if the given chord is 16 units in length.

11. Re: Problem with a chord

So, the number of chords of integral length (other than the diameter) that contain A is 22??