A semi circle, r=1, is bisected by a chord parallel to the base so the segment above the chord is...

[henrygreenall123]

I created a problem which I cannot solve. A semi circle, radius of 1 unit, has been bisected by a chord, parallel to the base, such that the segment above the chord is exactly half the area of the semi circle. At what height must this chord be drawn. Only wolfram alpha seems to be able to solve this? Even AI can’t. See the diagram for details. (Not drawn accurately)

 

[stapel]

I created a problem which I cannot solve. A semi circle, radius of 1 unit, has been bisected by a chord, parallel to the base, such that the segment above the chord is exactly half the area of the semi circle. At what height must this chord be drawn. Only wolfram alpha seems to be able to solve this? Even AI can’t. See the diagram for details. (Not drawn accurately) View attachment 36760

I'm pretty sure you need trigonometry for this. Have you studied trig yet?

 

[topsquark]

I created a problem which I cannot solve. A semi circle, radius of 1 unit, has been bisected by a chord, parallel to the base, such that the segment above the chord is exactly half the area of the semi circle. At what height must this chord be drawn. Only wolfram alpha seems to be able to solve this? Even AI can’t. See the diagram for details. (Not drawn accurately) View attachment 36760

That AI cannot solve this should come as no surprise. AI is not yet designed to "think" this way.

This is do-able, but you are going to wind up with a Trigonometric equation with a linear term that will need to be solved numerically.

The main key is to get the area of the bottom part of the 1/2 circle. That's two sectors and a triangle. Do you know how to find those areas?

-Dan

 

[henrygreenall123]

That AI cannot solve this should come as no surprise. AI is not yet designed to "think" this way.

This is do-able, but you are going to wind up with a Trigonometric equation with a linear term that will need to be solved numerically.

The main key is to get the area of the bottom part of the 1/2 circle. That's two sectors and a triangle. Do you know how to find those areas?

-Dan

Hi, thank you for the reply. The areas of the triangle and segments that you mentioned can only be calculated in terms of “k” (see diagram). Sorry but this doesn’t solve the question.

 

[henrygreenall123]

I'm pretty sure you need trigonometry for this. Have you studied trig yet?

I’ve studied a lot of trigonometry up to college level, I think the question is not solvable for a human being but I’d absolutely love someone to prove me wrong and get the correct answer without using wolfram etc.

 

[lev888]

I’ve studied a lot of trigonometry up to college level, I think the question is not solvable for a human being but I’d absolutely love someone to prove me wrong and get the correct answer without using wolfram etc.

The solution involves an equation that might look like this: x=sin(x)+1 (Look up the area of a segment of a circle formula). In general, such equations can only be solved using numerical methods.

 

[BigBeachBanana]

Putting the center at [imath](0,0)[/imath] and integrating over the bottom right area.
[math]\int_0^k \sqrt{1-y^2}\, dy = \dfrac{1}{2} \left(k\sqrt{1 - k^2} + \sin^{-1}(k)\right)=\dfrac{\pi}{8} \implies k \approx 0.403972753299517[/math]
[imath]k[/imath] was obtained via numerical approximation.

(Someone can fact-check the answer.)

 

[limiTS]

For a generic circle of radius [imath]R[/imath], the chord's height [imath]y[/imath] is described by this equation:

[math]\displaystyle R^2\left[\arcsin\left(\frac{\sqrt{R^2-y^2}}{R}\right)-\arcsin\left(\frac{y}{R}\right)\right]=2y\sqrt{R^2-y^2}[/math]
I used a root-finding technique to estimate [imath]y[/imath] when [imath]R=1[/imath]

Spoiler: Height

0.403972753299517

I'm not a mathematician, so maybe the others here can find a simpler relationship.

 

[blamocur]

Putting the center at [imath](0,0)[/imath] and integrating over the bottom right area.
[math]\int_0^k \sqrt{1-y^2}\, dy = \dfrac{1}{2} \left(k\sqrt{1 - k^2} + \sin^{-1}(k)\right)=\dfrac{\pi}{8} \implies k \approx 0.403972753299517[/math]
[imath]k[/imath] was obtained via numerical approximation.

(Someone can fact-check the answer.)

I checked your solution, the resulting area is [imath]\frac{\pi}{4}[/imath] with relative error of around [imath]10^{-15}[/imath]

 

[henrygreenall123]

I checked your solution, the resulting area is [imath]\frac{\pi}{4}[/imath] with relative error of around [imath]10^{-15}[/imath]

Yes it is correct but k is an irrational number and cannot be expressed by any fraction, therefore there will never be a 100% accurate answer sadly. I’ve always wondered if k has a direct link to Pi?

 

[topsquark]

Hi, thank you for the reply. The areas of the triangle and segments that you mentioned can only be calculated in terms of “k” (see diagram). Sorry but this doesn’t solve the question.

Or, alternately, [imath]\theta[/imath], with which you can then calculate the k. Did you even look at the method?

I wasn't planning on solving your question for you. I was giving you a hint on how to do it yourself.

-Dan

 

[blamocur]

Yes it is correct but k is an irrational number and cannot be expressed by any fraction, therefore there will never be a 100% accurate answer sadly.

Math is full of "inaccurate" numbers, like [imath]\sqrt{2}[/imath]. Just because we use special symbols like [imath]\pi[/imath] does not make the corresponding numbers "100% accurate". Not to get too philosophical, but you can use the text of the problem as a name for the number representing the solution. Just like I can use equation [imath]x^2-2=0[/imath] instead of [imath]\sqrt{2}[/imath] to define the same number.

 

[Dr.Peterson]

Yes it is correct but k is an irrational number and cannot be expressed by any fraction, therefore there will never be a 100% accurate answer sadly. I’ve always wondered if k has a direct link to Pi?

You seem to have been misled in your math education. Too often we teach only problems that can be solved exactly, and give the wrong impression that every problem you can make up can be solved in such a way. The reality is that most problems (especially real life problems) require numerical approximations like the methods that have been mentioned.

You need to change your expectations.

I’ve studied a lot of trigonometry up to college level, I think the question is not solvable for a human being but I’d absolutely love someone to prove me wrong and get the correct answer without using wolfram etc.

Rather, such questions can't be solved exactly by algebraic methods; but that doesn't mean they can't be solved. This is in fact true for most of trigonometry. Do you consider it a failure when you have to get a decimal answer by calculator and round it? That's actually a success.

 

[henrygreenall123]

Or, alternately, [imath]\theta[/imath], with which you can then calculate the k. Did you even look at the method?

I wasn't planning on solving your question for you. I was giving you a hint on how to do it yourself.

-Dan

I’ve been playing with this question for years. Theta is just as unsolvable as k. Doing it that way just gets you nowhere. You need theta to work out k and you need k to work out theta.

 

[pka]

I created a problem which I cannot solve. A semi circle, radius of 1 unit, has been bisected by a chord, parallel to the base, such that the segment above the chord is exactly half the area of the semi circle. At what height must this chord be drawn. Only wolfram alpha seems to be able to solve this? Even AI can’t. See the diagram for details. (Not drawn accurately) View attachment 36760

This is a most tiresome thread. Folks all it takes is basic high-school geometry.
We are given a semi-circle of radius [imath]R=1[/imath] hence the area of seni-circle [imath]A=\frac{1}{2}(\pi)(1)^2[/imath]
Thus the cord that is parallel to the diameter call it [imath]\mathcal{C}[/imath] its length is [imath]|\|\mathcal{C}\||= 2\sqrt{1-k^2}[/imath]
Where the value of the length [imath]}k[/imath] the question.
Here is a link in circular segments. On that diagram [imath]d[/imath] is our [imath]k[/imath]
.[imath][/imath][imath][/imath]

 

[Dr.Peterson]

Perhaps it is worth mentioning that this is a classic problem, perhaps first made famous as a Car Talk "puzzler" (originally in 2002, I believe). It is well known that the problem results in a transcendental equation that can only be solved numerically; but that equation can be obtained either by calculus or by geometry.

Here is one source I find:

Car Talk cylindrical fuel tank problem

This week’s Car Talk had a great problem from a caller (you can listen to that episode here – show #1045: Pi Over Two Dopes). In short, the guy calling drove an 18-wheeler truck. His fuel gauge was unreliable, so he used a dip-stick type method to measure the amount of fuel level. So, here […]

www.wired.com

More briefly,

Quarter-Tank Problem -- from Wolfram MathWorld

Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R...

mathworld.wolfram.com

 

[BigBeachBanana]

I’ve been playing with this question for years. Theta is just as unsolvable as k. Doing it that way just gets you nowhere. You need theta to work out k and you need k to work out theta.

I solved for k above in post#7. What do you mean by k is "unsolvable"?

 

[henrygreenall123]

I solved for k above in post#7. What do you mean by k is "unsolvable"?

How did you do it?

 

[BigBeachBanana]

How did you do it?

What did you not understand in post#7?

 

[stapel]

Perhaps it is worth mentioning that this is a classic problem, perhaps first made famous as a Car Talk "puzzler" (originally in 2002, I believe)....

I miss "Click & Clack, the Tappet Brothers". ?