## Posts tagged ‘triangle’

### Dos Equis XX Math Puzzles

No, the title of this post does not refer to the beer. Though it may be the most interesting blog post in the world.

It refers to the date, 10/10, which — at least this year — is the second day of National Metric Week. It would also be written in Roman numerals as X/X, hence the title of this post.

For today, I have not one, not *two*, but **three** puzzles for you. I’m providing them to you well in advance of October 10, though, in case you’re one of those clever types who wants to use these puzzles on the actual date… this will give you time to plan.

The first is a garden-variety math problem based on the date (including the year).

Today is 10/10/16. What is the area of a triangle whose three sides measure 10 cm, 10 cm, and 16 cm?

Hint:A triangle appearing in an analogous problem exactly four years ago would have had the same area.

The next two puzzles may be a little more fun for the less mathy among us — though I’m not sure that any such people read this blog.

Create a list of words, the first with 2 letters, the second with 3 letters, and so on, continuing as long as you can, where each word ends with the letter X. Scoring is triangular: Add the number of letters in all the words that you create until your first omission. For instance, if you got words with 2, 3, 4, 5, and 8 letters, then your score would be 2 + 3 + 4 + 5 = 14; you wouldn’t get credit for the 8-letter word since you hadn’t found any 6- or 7-letter words.

2 letters: _________________________

3 letters: _________________________

4 letters: _________________________

5 letters: _________________________

6 letters: _________________________

7 letters: _________________________

8 letters: _________________________

9 letters: _________________________

10 letters: _________________________

11 letters: _________________________

12 letters: _________________________

13 letters: _________________________

14 letters: _________________________

Note: There are answer blanks above for words up to 14 letters, because — you guessed it — the longest English word that ends with an X contains 14 letters.

The third and final puzzle is a variation on the second.

How many words can you think of that contain the letter X

twice? (Zoiks!) Scoring: Ten points for the first one, and a bazillion points for each one thereafter — this is hard! Good luck!

If you’re in desperate need of help, you can access **my list of words for both puzzles** — of which I’m fairly proud, since my list of words that end in X include math words for 2 through 10 letters — or do a search at www.morewords.com.

### Infinite Integer Triangles

Here’s an interesting question.

Given the side of a triangle with integer length, what is the set of all points in the plane for which the other two sides will also have integer lengths?

And by *interesting*, I mean that the answer wasn’t immediately obvious to me.

So I drew a segment 5 units long in Geometer’s SketchPad, created a bunch of concentric circles with integer radii and centers at the endpoints of the segment, identified the intersection points of those circles, and finally hid the circles. The result was the following beautiful image:

And by *beautiful*, I mean that the result is, well, beautiful. At least to a math dork. If this had been painted by Van Gogh, it would have been called **Triangle in a Starry Night**. (Okay, maybe not.)

The triangle indicated by the dashed lines is the famous 3-4-5 right triangle. The points in the upper right and upper left corners yield the less well known but similarly intoxicating 5-5-8 triangle. If the limitations of the web allowed this image to extend infinitely in all directions, the result would be infinite beauty. Alas, reality confines us.

I have an infinity of jokes that deal with triangles and circles, but I’ll only share a subset of them here.

What did the triangle say to the circle?

Your life is pointless.Why don’t circles hang out with ellipses?

Too eccentric.What did the hypotenuse say to the other two sides?

Nice legs!Where do circles and ellipses spend their vacations?

Coney Island.What’s a circle?

A round, straight line with a hole in the middle.What did the circle say to the tangent line?

Stop touching me!

### 6 Degrees of Bad Math Jokes

I once read an article that said, “To a greater or lesser degree, everything tastes like chicken.” Well, that’s true, but it’s also true that everything tastes like broccoli, **to a greater or lesser degree**. Carrots, to a greater degree; mint chocolate chip ice cream, to a lesser degree.

To a greater or lesser degree, some of the following jokes are funny.

What did the thermometer say to the graduated cylinder?

A scientist dropped a thermometer and a candle from the roof of a building. He observed that both objects reached the ground at the same time. Conclusion: A thermometer falls at the speed of light.

A doctor walks into a meeting, and a nurse asks why he has a rectal thermometer behind his ear. “Damn,” says the doctor, “some asshole has my pen!”

The star college football player was taking a math exam. The coach desperately needed him for the big game on Saturday, so the professor agreed to an oral exam.

“All right,” said the professor. “How many degrees are in a circle?”

“That depends,” said the boy. “How big is the circle?”

If you’re cold and there’s a right triangle nearby, stand in the corner opposite the hypotenuse. It’s always 90° over there.

The number you have dialed is imaginary. Please rotate your phone 90°, and try again.

### Dissections are Fun, No Matter How You Slice It

Some folks have referred to me as “a real cut‑up.” Of course, those folks are generally septuagenarians who still use words coined in the mid‑1800’s. But I take it as a compliment, perhaps because I’m a fan of dissection problems.

Speaking of dissection, here’s one of my favorite quotes about jokes, which comes from E. B. White:

Humor can be dissected as a frog can — but the thing dies in the process, and the innards are discouraging to any but the pure scientific mind.

Like jokes and amphibians, geometric figures can be dissected, too. The following are three of my favorite dissection problems.

My all-time favorite is the Haberdasher Problem, originally proposed by H. E. Dudeney in 1902. It’s my favorite because of its simplicity — it can be described with just a few words, and it asks the would‑be solver to turn the simplest of all polygons, an equilateral triangle, into a square:

With three cuts, dissect an equilateral triangle so that the pieces can be rearranged to form a square.

The second isn’t really a problem to be solved. Instead, it’s a dissection that offers a surprising result. (At least, I was surprised the first time I saw it.)

Remove a tetrahedron of edge length 1 unit from each vertex of a larger tetrahedron with edge length 2 units. In total, remove four tetrahedra, one from each vertex. The image below shows one of the tetrahedra being removed. What is the shape of the solid that remains?

Finally, here’s an original dissection problem.

Form a pentagon by placing an isosceles triangle with height 1 unit on top of a square with side length 2 units. Then find a point

Palong the perimeter of the pentagon such that whenPis connected to two vertices, the three pieces into which the pentagon is divided can be rearranged to form a square. Find the area of the largest piece.

As usual, these problems are presented without answers or solutions. More fun can be had if you solve them yourself.

### Fair and Square Solution

A few days ago, I posted the following problem:

Three points are randomly chosen along the perimeter of a square. What is the probability that the center of the square will be contained within the triangle formed by these three points?

In that post, I mentioned that I had an elegant solution. If you want to think about the problem a little before seeing my solution, check out the Three Points page at the MJ4MF website. I created a really cool Excel file that allows you to explore 100 cases at a time. I think it’s worth a look.

But if you’re not interested in any of that, here’s my solution…

Without loss of generality, choose any two points along the perimeter. Call them A and B. Then, construct segments AD and BC that pass through point P, the center of the square.

This gives four cases:

- If A and B are the two chosen points, then the third vertex must be chosen along segment CD to form a triangle that contains point P.
- If B and D are the two chosen points, then the third vertex must be chosen along segment CA.
- If D and C are the two chosen points, then the third vertex must be chosen along segment AB.
- If C and A are the two chosen points, then the third vertex must be chosen along segment BD.

What we’ve now done is selected two sets of points in **four** different ways, and the portions along the segments where the third point could be chosen collectively comprise the **entire perimeter**. Consequently, the probability that a point along the perimeter is 1, but this is divided among 4 cases, so the probability that a randomly selected triangle will contain the center is **1/4**.

As a footnote, points A and B were chosen arbitrarily, so there is no loss of generality.