[Prof. Templeton]

Professor Templeton liked to practice shooting basketballs in his spare time. He wanted to convert his courtyard into a semicircular 3 point shooting court. He enlisted the help of fellow Braindenizen Bonanova to help him lay out the semicircle into his 1 unit by 1 unit courtyard. When Bonanova arrived Prof. T showed him his square courtyard and eager to show off his math skills he said, "If we lay out the semicircle inside the square, it will lay inside half the square and it's radius will be half that of one side of the square courtyard so it will have an area of

1/2*pi*(1/2)² or .3927". Bonanova shook his head and smirked at the Prof. "You want to make the semicircle as large as possible don't you?", he replied. "Um, of course", said Prof. T, "isn't that the largest semicircle we can make, it will be tangent with three sides of the square after all". "Not even close", commented Bonanova, and he proceeded to lay out the largest semicircle that would fit inside the 1 unit by 1 unit courtyard. What was the area of Prof. Templeton's new semicircular 3 point basketball court?

[Guest]

Professor Templeton liked to practice shooting basketballs in his spare time. He wanted to convert his courtyard into a semicircular 3 point shooting court. He enlisted the help of fellow Braindenizen Bonanova to help him lay out the semicircle into his 1 unit by 1 unit courtyard. When Bonanova arrived Prof. T showed him his square courtyard and eager to show off his math skills he said, "If we lay out the semicircle inside the square, it will lay inside half the square and it's radius will be half that of one side of the square courtyard so it will have an area of

1/2*pi*(1/2)² or .3927". Bonanova shook his head and smirked at the Prof. "You want to make the semicircle as large as possible don't you?", he replied. "Um, of course", said Prof. T, "isn't that the largest semicircle we can make, it will be tangent with three sides of the square after all". "Not even close", commented Bonanova, and he proceeded to lay out the largest semicircle that would fit inside the 1 unit by 1 unit courtyard. What was the area of Prof. Templeton's new semicircular 3 point basketball court?

.539

semicircle must be tangent to 2 sides and the end points must touch the other 2 sides

Edited December 5, 2008 by HokieKen

[Prof. Templeton]

.539

semicircle must be tangent to 2 sides and the end points must touch the other 2 sides

Excellent picture. Can you explain your answer further?

[Guest]

Excellent picture. Can you explain your answer further?

Actually, no. It's just geometrically intuitive. Once I saw the requirement, I drew it and measured and did calculations. I can't lay the geometry or the trig out because I'm not entirely sure how I knew it

[Guest]

Actually, no. It's just geometrically intuitive. Once I saw the requirement, I drew it and measured and did calculations. I can't lay the geometry or the trig out because I'm not entirely sure how I knew it

Really not being rude here, but it's just like a professor to want to know how you got your answer.

[Yoruichi-san]

Okay...after much struggling with the drawing tool...(I'm totally illiterate when it comes to that stuff XP)...

The semi-circle with radius R is half of the blue circle, which is inscribed in the larger black square of side 2R, and inscribes the smaller dashed square of diagonal 2R, and hence side 2^1/2R. The red square is the yard of side 1, and since it is shorter than the large square by the same distance it is longer than the small square, the side of the red square is the average of the sides of the two squares.

This gives us: (2R+2^1/2R)/2=1 -> R=1/(1-1/2^1/2)=0.585789

So the area of the semi-circle is piR²/2=0.539012

[bonanova]

Consider rotating the semicircle occupying half the square ccw through an angle a:

The courtyard is shown in red.

As the angle a increases from 0, the height of the inscribed semicircle increases from 1 to 1+d.

When a increases beyond 45^o, the width of the circle begins to decrease.

That establishes 45^o as the optimum angle.

The height of the courtyard is r + r cos[a] = 1.

At 45^o, r = .5858 units and the shaded area = .539 square units.

[Guest]

Let me explain:

There is a equilateral triangle in the figure, and A is a side of it.

The hypotenous is also diameter of circle = A.2^1/2

Then Radius= A/2^1/2

If you look at the small triangle, its hypotenous is radius, then sides of it is=A/2

Now if you look at the lower line of square, the A length is divided to two A/2 parts, at B.

From B point to lower right corner of square = A/2 + x.

This length is also a radius. (look at O point to right side of square)

Then Radius= A/2 + x

Also Radius was = A/2^1/2

This statement gives x=A(2^1/2-1)/2

The lower side of square shows that A + x = 1 (unit of square)

If you solve this statement, the result is=0.5387, just as Hokieken found, consciously or not.

I've skipped some operations for not to make longer my post. But as the problem is not so hard, no need for them?

[Prof. Templeton]

Good job all! Using HK original drawing as a guide, if the whole circle is allowed to go outside the square, what is it's relationship with the square's vertex A.

[Guest]

Good job all! Using HK original drawing as a guide, if the whole circle is allowed to go outside the square, what is it's relationship with the square's vertex A.

A is a point on the circle. The bottom corner of square can be divided into 2 equilateral triangles in which all sides must be equal to the radius of the circle because it is oriented at 45 deg. Therefore the distance from the center of the circle to A = r.

Edited December 5, 2008 by HokieKen

[Guest]

Here is sketch

..

[Yoruichi-san]

Good job all! Using HK original drawing as a guide, if the whole circle is allowed to go outside the square, what is it's relationship with the square's vertex A.

Or you could just look at my drawing ;P

[Prof. Templeton]

Or you could just look at my drawing ;P

You said yourself your not to handy with those things. ;P® Besides, it looks like your circle doesn't quite get there, but you know being a Prof. I'm looking for a good proof.

[Yoruichi-san]

You said yourself your not to handy with those things. ;P® Besides, it looks like your circle doesn't quite get there, but you know being a Prof. I'm looking for a good proof.

Yeah...stupid drawing tool only lets me resize the square by certain amounts...it won't put the vertex exactly on the circle... <_<

It was more the method, which was what the picture was illustrating than the actual picture, which I tried (and I guess must have failed

) to explain...

Since the diameter of the circle intersects your square, and the triangle formed has A as a right angle (by definition of a square), then A must be an inscribed angle, hence on the circumference of the circle. Put simply...the angle A subtends the diameter.