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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

What is the largest square that can be inscribed in a 3-4-5 triangle?

2 + 2 = 5, for large values of 2.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

Well, we could start off by making a little formula to figure out the area at any point.

Let us have the triangle like this:

A

.

|\

| \

| \

|____\

B C

I imagine that the square would have one corner at **B** and another corner somewhere between **A** and **C**

If we drew the triangle on some graph paper, we could have the lower corner **B** at (0,0), the upper corner **A** at (0,4) and the far corner **C** at (3,0).

Ummm ....

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

As an extension, find the largest (in terms of area) such *rectangle*. Generalise to an a-b-c right triangle.

2 + 2 = 5, for large values of 2.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

First I was wary about helping, because as you used phrases such as 'as an extension' and 'generalise', it looked suspiciously like something that you had to do for school. However, due to a combination of the facts that it has now been quite a long time, so if it was work you would most probably have handed it in by now, and you gave that cool trigonometry help thingy on the 'This is Cool' board, and it was such an interesting puzzle, I decided to help.

Thinking about the corner opposite the corner of the square that is also in the corner of the triangle, we know that it must have the same x and y co-ordinates (because it's a square) and that the point must lie somewhere on the diagonal line of the triangle (because it's a big square). Using MathsIsFun's graph idea, the equation of the diagonal line in the triangle would be y=-4/3x+4. We also know that the square's opposite corner must lie somewhere on y=x, so we have 2 simultaneous equations.

Substitute in y=x and you get x=-4/3x+4.

Add 4/3x: 7/3x=4

Divide by 7/3: x=4/ (7/3)=12/7.

This means that the largest square has a length of 12/7 units and an area of 144/49, or 2 44/49, square units.

Unfortunately, this post is in the realm of the chocolate teapot until someone proves that the biggest square must have the same corner as the triangle and I'm terrible at proofs. The pigeon-hole principle might do it, but I'm not exactly sure how to use it on this.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

For the extension, I used advanced calculus that I won't bore you with now (unless someone requests it), but basically, if the triangle has perpendicular sides of lengths a and b, then the biggest rectangle will always have sides of length a/2 and b/2, subject to the proof being proved that I mentioned previously. This obviously means that the rectangle will have an area of ab/4, which means that its area will also be half of that of the triangle.

Why did the vector cross the road?

It wanted to be normal.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

I'm not a student -- at least not in the formal sense -- so no worries there! I just posted it as an interesting puzzle in its own right, and subsequently thought of the generalisation to a rectangle.

The rectangle problem can be solved without calculus. (Anyway, did you really mean to imply that advanced calculus is boring?!)

In your diagram above, if the square has a vertex at (0,0) and side a, by similar triangles, (4-a)/a = 4/3, so a = 12/7.

If the square has one side on the hypotenuse, the small triangle at the top is similar to the whole 3-4-5 triangle. By area considerations, the height of the 3-4-5 triangle is 12/5. So, by similar triangles, a/5 = (12/5 - a)/(12/5), from which a = 60/37; a tad smaller than 12/7 = 60/35.

If the square is to be inscribed, i.e., each vertex touches the triangle, these are the only two possibilities.

For the inscribed rectangle, suppose the rectangle has vertices (0,0), (0,r), and (s,0), and the triangle has vertices (0,a) and (b,0). Then by similar triangles, as above, a/b = (a - r)/s, so s = b - (b/a)r. Then the area of the rectangle, A = rs = br - (b/a)r² = (b/a)(ar - r²). Completing the square, A = (b/a)[(a²/4) - (r - a/2)²]. So the maximum area of ab/4 occurs when r = a/2, as you said.

The case where the rectangle has one side on the hypotenuse may be handled similarly. Is the maximum area in that case <. =, or > ab/4? I won't spoil the fun by giving that away!

2 + 2 = 5, for large values of 2.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

Many ways to skin a cat, hey?

Yes, I started off with the "equation of line" approach, because ... I don't know why ... I guess I was just teasing the problem apart. I was kinda looking forward to optimising the area of that rectangle problem in Excel - to see if it had a pretty curve, and whether the answer was some magic number. Maybe I will one day.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

1) Sorry for thinking that you were trying to cheat.

2) Of course advanced calculus isn't boring! One of the reasons why I used it!

3) Um, thats it, really.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

NIH wrote:

The case where the rectangle has one side on the hypotenuse may be handled similarly. Is the maximum area in that case <. =, or > ab/4? I won't spoil the fun by giving that away!

Nor will I. It's annoying how after you do lots of complicated work to find the answer, you can tell just by looking though!

*Last edited by mathsyperson (2005-06-26 08:55:48)*

Why did the vector cross the road?

It wanted to be normal.

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