## Poll

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**8 members have voted**

February 19th, 2020 at 7:40:39 AM
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In our continued celebration of 2020, the year of math appreciation, I present the following geometry problem.

The following diagram shows a chord drawn across a circle. From a point on the chord, a perpendicular line is drawn to another point on the same circle. Distances are shown in the diagram above.

What is the radius of the circle?

Usual rules:

1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.

2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. This rule has been violated by past winners the previous two problems. Past winners who must chime in early, may PM me.

3. Beer to the first satisfactory answer and solution, subject to rule 2.

4. Please put answers and solutions in spoiler tags.

The following diagram shows a chord drawn across a circle. From a point on the chord, a perpendicular line is drawn to another point on the same circle. Distances are shown in the diagram above.

What is the radius of the circle?

Usual rules:

1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.

2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. This rule has been violated by past winners the previous two problems. Past winners who must chime in early, may PM me.

3. Beer to the first satisfactory answer and solution, subject to rule 2.

4. Please put answers and solutions in spoiler tags.

It's not whether you win or lose; it's whether or not you had a good bet.

February 19th, 2020 at 1:49:13 PM
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Is the question being asked clear?

It's not whether you win or lose; it's whether or not you had a good bet.

February 19th, 2020 at 1:58:17 PM
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Crystal.Quote:WizardIs the question being asked clear?

It looks like it should be easy.... But, I'm getting in a right mess trying :o)

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

February 19th, 2020 at 2:51:19 PM
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Placeholder while I paste my workings 5.409771

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

February 19th, 2020 at 2:59:35 PM
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Quote:OnceDearNot sure if I'm a beer recipient, so workings PM'd to Wizard.Placeholder while I paste my workings 5.409771

Negative. No beer for you yet. Since you're on the wagon, I hope a spotted dick will suffice.

It's not whether you win or lose; it's whether or not you had a good bet.

February 19th, 2020 at 3:06:28 PM
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Off to check my algebra.....

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

February 19th, 2020 at 3:28:29 PM
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Quote:OnceDearOff to check my algebra.....

I had messed up in algebra. Here's my solution after some checking and fixing..

Label the 3 points ABC from left to right.

AC=7.

The line AC is distance z vertically from the circle's centre

Half way along AC is at 3.5 and is above the origin of the circle.

For A, it's offset 3.5 to the left of the centre of the circle, so x1= -3.5 and y1=z

For B, it's offset 0.5 to the left of the centre of the circle so x2= -0.5 and y2=z+2

For C, it's offset 3.5 to the right of the centre of the circle, so x3= 3.5 and y3=z

x1^2 +y1^2 =x2^2 +y2^2 = x3^2 +y3^2 =r^2 ( Pythagoras applied to any Right angle triangle made from centre to point xy on the circle of radius r where x and y coordinates from the centre)

(-3.5)^2 + z^2 = (-0.5)^2 +(z+2)^2

(-3.5)^2 + z^2 = (-0.5)^2 + z^2 +4z +4

(-3.5)^2 +z^2 -(-0.5)^2 -z^2 -4z -4 = 0

z^2 - z^2 - 4z = (-0.5)^2-(-3.5)^2 +4

-4z = -8

z= 2

now we know z

x1^2 +y1^2=r^2

(-3.5)^2+z^2=r^2

(-3.5)^2+2^2=r^2

r^2=16.25

r=4.03112887415

AC=7.

The line AC is distance z vertically from the circle's centre

Half way along AC is at 3.5 and is above the origin of the circle.

For A, it's offset 3.5 to the left of the centre of the circle, so x1= -3.5 and y1=z

For B, it's offset 0.5 to the left of the centre of the circle so x2= -0.5 and y2=z+2

For C, it's offset 3.5 to the right of the centre of the circle, so x3= 3.5 and y3=z

x1^2 +y1^2 =x2^2 +y2^2 = x3^2 +y3^2 =r^2 ( Pythagoras applied to any Right angle triangle made from centre to point xy on the circle of radius r where x and y coordinates from the centre)

(-3.5)^2 + z^2 = (-0.5)^2 +(z+2)^2

(-3.5)^2 + z^2 = (-0.5)^2 + z^2 +4z +4

(-3.5)^2 +z^2 -(-0.5)^2 -z^2 -4z -4 = 0

z^2 - z^2 - 4z = (-0.5)^2-(-3.5)^2 +4

-4z = -8

z= 2

now we know z

x1^2 +y1^2=r^2

(-3.5)^2+z^2=r^2

(-3.5)^2+2^2=r^2

r^2=16.25

r=4.03112887415

February 19th, 2020 at 3:48:04 PM
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Two coordinates will solve the problem, (-0.5, 2+y) and (3.5, y)

(-0.5)^2+ (2+y)^2 = r^2..........A

(3.5)^2 + y^2 = r^2...................B

A - B :-

4y+4 =12 = 0

y=2, r = (65)^0.5/2

(-0.5)^2+ (2+y)^2 = r^2..........A

(3.5)^2 + y^2 = r^2...................B

A - B :-

4y+4 =12 = 0

y=2, r = (65)^0.5/2

Last edited by: ssho88 on Feb 19, 2020

February 19th, 2020 at 3:50:20 PM
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We concur, though your calcs seem simpler than mine. Your last line ( the big number?) seems a bit of a leap to my addled brain.Quote:ssho88Two coordinates will solve the problem, (-0.5, 2+y) and (3.5, y)

(-0.5)^2+ (2+y)^2 = r^2..........A

(3.5)^2 + y^2 = r^2...................B

A - B :-

4y+4 =12 = 0

y=2, r^2 = (65)^0.5/2

February 19th, 2020 at 4:24:12 PM
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Quote:OnceDearI had messed up in algebra. Here's my solution after some checking and fixing..

Label the 3 points ABC from left to right.

AC=7.

The line AC is distance z vertically from the circle's centre

Half way along AC is at 3.5 and is above the origin of the circle.

For A, it's offset 3.5 to the left of the centre of the circle, so x1= -3.5 and y1=z

For B, it's offset 0.5 to the left of the centre of the circle so x2= -0.5 and y2=z+2

For C, it's offset 3.5 to the right of the centre of the circle, so x3= 3.5 and y3=z

x1^2 +y1^2 =x2^2 +y2^2 = x3^2 +y3^2 =r^2 ( Pythagoras applied to any Right angle triangle made from centre to point xy on the circle of radius r where x and y coordinates from the centre)

(-3.5)^2 + z^2 = (-0.5)^2 +(z+2)^2

(-3.5)^2 + z^2 = (-0.5)^2 + z^2 +4z +4

(-3.5)^2 +z^2 -(-0.5)^2 -z^2 -4z -4 = 0

z^2 - z^2 - 4z = (-0.5)^2-(-3.5)^2 +4

-4z = -8

z= 2

now we know z

x1^2 +y1^2=r^2

(-3.5)^2+z^2=r^2

(-3.5)^2+2^2=r^2

r^2=16.25

r=4.03112887415

I agree!!! Congratulations, you have finally earned a beer. Substitutions for people on the wagon are gladly accommodated. Hope to pay off my debt soon.

The formula in your solution I did not know. My solution is quite a bit longer, but requires knowing only the Pythagorean formula and it is applied once only.

It's not whether you win or lose; it's whether or not you had a good bet.