Semicircle Chord Cut v2 - Solutions & Notes

To mark my 150th Problem of the Day, at the end July 2017, I decided to try and compose a problem that was a true problem for myself.  That is, a question for which I did not know the answer and not immediately aware of an effective solution strategy ... and then have some fun trying to solve it.  I did have fun ... and came up with two different ways to solve it - and was also delighted that the solution contained a very interesting and unexpected feature.

The problem that I devised was motivated by a problem that I found in a short article entitled How Do You Slice the Bread?  The authors of the article model a slice of bread in two dimensions with a rectangle and a semi-ellipse (as shown in diagram).  One of the questions they pose – and solve – is where to cut the slice of bread to produce two “triangular” pieces that are equal in area such that the ‘cut’ must start from one of the corners of the lower rectangular section.

This helped me come up of with a similar question ...

Where do you cut a semicircle – starting from one of the endoints of the diameter – to create two regions of equal area?

Although someone may have posed and solved this problem previously, I’m not aware of such and was surprised not to find anything after quite a bit of searching on the internet.  My work on solving the problem required me to put on my problem-solving hat and led to some engaging mathematics and a solution with an interesting and unexpected property.  It also led me to two different approaches to the solution and to apply a mix of mathematics software to assist in the solution and in a visual verification of it.

Two different solution strategies

Given the formulas utilized in my two solution strategies, I will refer to the first strategy I developed as the integration strategy (calculus approach), and the alternative strategy as the area of sector strategy (non-calculus approach).  Both strategies started with the following set-up and diagram.

Set-up

The graph of $$f\left( x \right) = \sqrt {{k^2} - {x^2}}$$ is a semicircle with x-intercepts of $$\left( { - k,\;0} \right)$$ and $$\left( {k,\;0} \right)$$ labelled A and B in the diagram.  A line segment (chord) is drawn from A to a point P on the circle that has coordinates $$\left( {p,\;\sqrt {{k^2} - {p^2}} } \right)$$
Objective:
Find the value of p, expressed in terms of k, so that the line segment (chord) AP divides the semicircle into two regions with equal area.

Integration strategy (calculus approach)

I begin by dividing the lower (blue) region into two parts – one is a right triangle with vertices A, P & D (added to diagram left).

The area of the semicircle$$= \frac{{{\rm{\pi }}{k^2}}}{2}$$; thus, each of the two regions (yellow & blue) need to have an area of $$\frac{{{\rm{\pi }}{k^2}}}{4}$$.

area of right $$\Delta$$APD$$= \frac{{k + p}}{2}\sqrt {{k^2} - {p^2}}$$

area of region bounded by DB, arc BP and PD is equal to the definite integral $$\int_p^k {\sqrt {{k^2} - {x^2}} \,dx}$$

Now, I use integration by parts $$\left( {\int {u\,dv} = uv - \int {v\,du} } \right)$$ to find $$\int {\sqrt {{k^2} - {x^2}} \,dx}$$.

let $$u = \sqrt {{k^2} - {x^2}}$$ and $$dv = dx\;\;\;\;\; \Rightarrow \;\;\;\;\;du = \frac{d}{{dx}}\left( {\sqrt {{k^2} - {x^2}} } \right)$$  and  $$v = \int {dx = x}$$

need to find $$\frac{d}{{dx}}\left( {\sqrt {{k^2} - {x^2}} } \right)$$

$$\frac{d}{{dx}}\left( {{{\left( {{k^2} - {x^2}} \right)}^{\frac{1}{2}}}} \right) = \frac{1}{2}{\left( {{k^2} - {x^2}} \right)^{ - \,\frac{1}{2}}}\left( { - 2x} \right) = \frac{{ - x}}{{\sqrt {{k^2} - {x^2}} }}\;\;\; \Rightarrow \;\;\;u = \frac{{ - x}}{{\sqrt {{k^2} - {x^2}} }}$$

substituting gives:  $$\int {\sqrt {{k^2} - {x^2}} \,dx} = x\sqrt {{k^2} - {x^2}} - \int {\left( {\frac{{ - {x^2}}}{{\sqrt {{k^2} - {x^2}} }}} \right)} \,dx = x\sqrt {{k^2} - {x^2}} + \int {\frac{{{x^2}}}{{\sqrt {{k^2} - {x^2}} }}} \,dx$$

$$= x\sqrt {{k^2} - {x^2}} + \int {\frac{{{k^2} - {k^2} + {x^2}}}{{\sqrt {{k^2} - {x^2}} }}} \,dx = x\sqrt {{k^2} - {x^2}} + \int {\frac{{{k^2}}}{{\sqrt {{k^2} - {x^2}} }}} \,dx - \int {\frac{{{k^2} - {x^2}}}{{\sqrt {{k^2} - {x^2}} }}} \,dx$$

$$= x\sqrt {{k^2} - {x^2}} + {k^2}\int {\frac{1}{{\sqrt {{k^2} - {x^2}} }}} \,dx - \int {\sqrt {{k^2} - {x^2}} } \,dx$$

$$\int {\sqrt {{k^2} - {x^2}} \,dx} = x\sqrt {{k^2} - {x^2}} + {k^2}\arcsin \left( {\frac{x}{k}} \right) - \int {\sqrt {{k^2} - {x^2}} } \,dx$$   $$\left[ {{\rm{add }}\int {\sqrt {{k^2} - {x^2}} } \,dx\;{\rm{to}}\;{\rm{both}}\;{\rm{sides}}} \right]$$

$$2\int {\sqrt {{k^2} - {x^2}} \,dx} = x\sqrt {{k^2} - {x^2}} + {k^2}\arcsin \left( {\frac{x}{k}} \right)\;\;\; \Rightarrow \;\;\;\int {\sqrt {{k^2} - {x^2}} \,dx} = \frac{x}{2}\sqrt {{k^2} - {x^2}} + \frac{{{k^2}}}{2}\arcsin \left( {\frac{x}{k}} \right)$$

$$\int_p^k {\sqrt {{k^2} - {x^2}} \,dx} = \left[ {\frac{x}{2}\sqrt {{k^2} - {x^2}} + \frac{{{k^2}}}{2}\arcsin \left( {\frac{x}{k}} \right)} \right]_p^k =$$

$$= \left( {0 + \frac{{{k^2}}}{2}\arcsin \left( 1 \right)} \right) - \left( {\frac{p}{2}\sqrt {{k^2} - {p^2}} + \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right)} \right) =$$

$$= \frac{{{\rm{\pi }}{k^2}}}{4} - \frac{p}{2}\sqrt {{k^2} - {p^2}} - \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right)$$

For the areas of the two regions to be equal, the area of the lower blue region needs to be $$\frac{{{\rm{\pi }}{k^2}}}{4}$$.

Thus, $$\frac{{{\rm{\pi }}{k^2}}}{4} - \frac{p}{2}\sqrt {{k^2} - {p^2}} - \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right) + \frac{{k + p}}{2}\sqrt {{k^2} - {p^2}} = \frac{{{\rm{\pi }}{k^2}}}{4}$$

$$\frac{p}{2}\sqrt {{k^2} - {p^2}} + \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right) - \frac{{k + p}}{2}\sqrt {{k^2} - {p^2}} = 0$$   $$\left[ {{\rm{subtracted}}\;\frac{{{\rm{\pi }}{k^2}}}{4}\;{\rm{to}}\;{\rm{both}}\;{\rm{sides}}} \right]$$

$$\frac{k}{2}\sqrt {{k^2} - {p^2}} = \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right)$$

Thus, the relationship between p and k is $$\sqrt {{k^2} - {p^2}} = k\arcsin \left( {\frac{p}{k}} \right)$$.

Before I explore this relationship between p and k further, let's see another way to achieve the same result.

Area of sector strategy (non-calculus approach)

There is another way to approach this problem that does not involve calculus, but I did not realize that until after I had used the previous strategy. If I was to give a hint to someone who was having difficulty coming up with a solution strategy, I would nudge them in the direction of this strategy. It's a bit more direct, has less computation, and does not require the application of integral calculus. Makes me wonder why I did not devise this strategy first.

Rather than finding the area of the lower region, I find the area of the upper (yellow) region which is a segment of the semicircle.  This is done by subtracting the area of triangle AOP from the area of sector AOP (shaded with hatched lines in diagram).

I first find an expression for angle AOP.  Angle BOP, labelled $${\rm{\theta }}$$ in the diagram, is $$\arccos \left( {\frac{p}{k}} \right)$$.  Thus, angle AOP$$= {\rm{\pi }} - {\rm{\theta }} = {\rm{\pi }} - \arccos \left( {\frac{p}{k}} \right)$$.

Using the formula for area of a sector $$A = \frac{1}{2}{\rm{\theta }}{r^2}$$, and the formula for area of a triangle $$A = \frac{1}{2}ab\sin C$$:

area of sector AOP $$= \frac{1}{2}\left( {{\rm{\pi }} - \arccos \left( {\frac{p}{k}} \right)} \right){k^2}$$; and, area of triangle AOP $$= \frac{1}{2}{k^2}\sin \left( {{\rm{AOP}}} \right)$$

note$$\sin {\rm{\theta }} = \sin \left( {{\rm{\pi }} - {\rm{\theta }}} \right)$$; thus, $$\sin {\rm{\theta }} = \sin \left( {{\rm{AOP}}} \right)$$ and $$\sin {\rm{\theta }} = \frac{{\sqrt {{k^2} - {p^2}} }}{k}$$

Thus, the area of triangle AOP$$= \frac{1}{2}{k^2}\sin {\rm{\theta }} = \frac{{{k^2}}}{2} \cdot \frac{{\sqrt {{k^2} - {p^2}} }}{k} = \frac{{k\sqrt {{k^2} - {p^2}} }}{2}$$

Then, area of upper yellow region (segment)$$= \frac{1}{2}\left( {{\rm{\pi }} - \arccos \left( {\frac{p}{k}} \right)} \right){k^2} - \frac{{k\sqrt {{k^2} - {p^2}} }}{2}$$

Since the area of the semicircle is $$\frac{{{\rm{\pi }}{k^2}}}{2}$$, then the area of upper yellow region is $$\frac{{{\rm{\pi }}{k^2}}}{4}$$.

$$\frac{1}{2}\left( {{\rm{\pi }} - \arccos \left( {\frac{p}{k}} \right)} \right){k^2} - \frac{{k\sqrt {{k^2} - {p^2}} }}{2} = \frac{{{\rm{\pi }}{k^2}}}{4}$$

$$\frac{1}{2}\left( {\frac{{\rm{\pi }}}{2} + \frac{{\rm{\pi }}}{2} - \arccos \left( {\frac{p}{k}} \right)} \right){k^2} - \frac{{k\sqrt {{k^2} - {p^2}} }}{2} = \frac{{{\rm{\pi }}{k^2}}}{4}$$     note: $$\frac{{\rm{\pi }}}{2} - \arccos \left( {\frac{p}{k}} \right) = \arcsin \left( {\frac{p}{k}} \right)$$

$$\frac{1}{2}\left( {\frac{{\rm{\pi }}}{2} + \arcsin \left( {\frac{p}{k}} \right)} \right){k^2} - \frac{{k\sqrt {{k^2} - {p^2}} }}{2} = \frac{{{\rm{\pi }}{k^2}}}{4}$$

$$\frac{{{\rm{\pi }}{k^2}}}{4} + \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right) - \frac{{k\sqrt {{k^2} - {p^2}} }}{2} = \frac{{{\rm{\pi }}{k^2}}}{4}$$

$$\frac{{k\sqrt {{k^2} - {p^2}} }}{2} = \frac{{{k^2}}}{2}\arcsin \left( {\frac{p}{k}} \right)$$

Thus, the relationship between p and k is:  $$\sqrt {{k^2} - {p^2}} = k\arcsin \left( {\frac{p}{k}} \right)$$.

The Result - Comments

Recall that the original objective of this problem was to find the value of p, expressed in terms of k, so that the line segment (chord) AP divides the semicircle into two regions with equal area.  That is, find the x-coordinate of one endpoint of the chord that cuts the semicircle into two regions of equal area when the other endpoint is $$\left( { - k,\;0} \right)$$ and the center of the 'orignal' circle is at $$\left( {0,\;0} \right)$$.

The working from both approaches shown above did not produce an answer where p is expressed explicitly in terms of k.  The result $$\sqrt {{k^2} - {p^2}} = k\arcsin \left( {\frac{p}{k}} \right)$$ does give the relationship between p and k, so that one could substitute in a value of k and then compute the value of p. Of course, all semicircles are geometrically similar which means that there will be a direct relationship between p and k. Hence, it makes sense to solve for p when $$k = 1$$ .  This particular value of p, call it a, can then be used as a constant to express p explicitly in terms of k.  How does one solve for p when $$k = 1$$ ?  Well, it's clear that technology will be required - and it proved very interesting (nearly as interesting as coming up with the two solution strategies) to explore the use of different math technology tools to find a.  I will produce another page soon that will demonstrate - and comment on - the different technological approaches to solving for a.

It turns out that $$a \approx 0.673612029183215...$$   therefore, $$p \approx 0.67361k$$

This number a is kind of 'special'.  Remember, a is the value of p when $$k = 1$$ in the equation $$\sqrt {{k^2} - {p^2}} = k\arcsin \left( {\frac{p}{k}} \right)$$. Thus, a is the number that solves the equation $$\sqrt {1 - {p^2}} = \arcsin \left( p \right)$$.  This has a clear geometric interpretation.  Consider the same equation in terms of x$$\sqrt {1 - {x^2}} = \arcsin \left( x \right)$$.  The value of x that solves this equation will be the x-coordinate of a point, call it P, on the unit circle such that the y-coordinate of point P $$\left[ {\sqrt {1 - {x^2}} } \right]$$ is equal to the angle (in radians) whose terminal ray goes through P.  And since it's the unit circle the length of the arc on the unit circle cut out by this angle will be equal to the vertical distance to point P.  This is a unique point on the unit circle - and it's where we make the 'cut' to divide the 'upper' half (semicircle) of the unit circle to divide it into two pieces having equal areas.  It definitely helps to see a visual (and dynamic) illustration.  Take a look at my Geogebra applet this is on the following page: semicircle chord cut v2 (applet)

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