Two weeks ago on car talk, Click and Clack heard an interesting gas tank turned geometry problem that was simple to state, but not so easy to solve. One of their callers had a cylindrical gas tank in his truck, and because the gas gauge was broken, this caller would drop a stick into the tank and check the gas level. The question was where he needed to mark his stick so that he could properly measure a quarter tank of gas.

Now if you’re thinking, that’s easy, it’s just a quarter of the height of the stick, then a picture will make you feel less sure:

The key is that the tank is not standing upright like a soda can, but is instead laying on its side, like a toilet paper roll. So let’s draw the cylindrical tank laying on its side, as the caller described. Then draw a stick that’s stuck in there vertically. If the tank is 20 inches in diameter, then a 20 inch stick will go from top to bottom, and 10 inches will surely be the ½ full mark. But where is the ¼ full mark?

Well, Click and Clack rightfully turned the problem into a 2-dimensional one, by rotating the tank around so that all you could see was the circular base. That turns the volume problem into a circle area problem, which will make life easier. Then, as they often do, they simplified the problem by splitting the circle in half, and focused on solving for the height at which the semicircle is split into two.

That’s called progress! But then they got stuck. They started talking about how you needed calculus to solve this problem, and then suggested that the caller find someone with a tank that is a quarter full, and use that to calibrate the stick

In response to what must have been overwhelming fan-mail, this past week Tom and Ray returned to the problem, and went over some interesting insights that a variety of their listeners and family members had shared. Yes, calculus will help to solve the problem (Wired does a nice analysis). However, trigonometry is really all that is needed (Dr. Math to the rescue!).

Here at Virtual Nerd we love to see problems like this generate so much discussion, and we can’t help but throw in our 2 cents. As you’ll see below, we’ve got an approach that gets pretty darn close, and doesn’t require any trig at all!

Let’s redraw the diagram and label and shade some things, like the radius (r), the height (h) from the center of the circle to the level that splits the semicircle in half, and the two equal regions A and B that we get after we split the semicircle:

We know that the area of our semicircle is ½ of the area of the whole circle. The whole circle has area πr², and so our semicircle has area ½πr². Now the shaded regions A and B both are supposed to have half the area of this semicircle. We can write that in math terms, by writing B=½•(½πr²). That is one equation that we have, but so what?

Well now we need to find the area of that shaded region B, and that is tough, because it’s a weird shape. There is a way to find the exact area using some trigonometry, but what if we did something silly. What if we stepped back and said, you know what, that area kind of looks like a rectangle!

Yeah, it kind of does. It’s a rectangle with a height of h, and a base of 2r, and the area of a rectangle is base times height. That means that our approximate value for the area of region B is B=2rh.

Now something magical happens! We can set our two equations for B equal to each other:

To solve this equation, we want to get h by itself, but we’ve got a lot of stuff going on. It turns out that there are many ways to proceed, but let’s just try to make life easier by considering the numbers first, and dealing with the variables second. We’ve got 2 on the left hand side. It would be nice to get it to the right side, away from the h. Well we can divide both sides by 2, or multiply both sides by ½ it’s really comes down to whatever you prefer. Multiplying will work nicely, so lets do that.

½•2 cancels out on the left side, and now the right side has ½•½ and then another ½ in the parentheses. Since we’re multiplying all the way through, it’s safe to drop the parentheses and simply evaluate ½•½•½, which equals 1/8.

Now we have h almost by itself, except there is a pesky r still with us on the left side. Well, that’s ok, because we can just divide both sides by r!

That leaves h on the left side, and since r²/r = r, we are left with (1/8)•π•r on the right side.

So, if our radius is 10 inches, we take out our calculator, type in 1/8•3.14159•10, and we get h= 3.93 inches. That means that if we mark the stick 3.93 inches from the bottom, we should have a pretty good measure of 1/4 tank of gas. Now that’s pretty cool, because the exact answer to this problem, as others have found and explained, is about 4.04 inches!

This is a great example of how simple estimation can go a long way to solving a complicated problem. Sure region B is not a rectangle, but it’s close, and so a rectangular approximation works pretty well. Plus, we know that region B is a little less than the area of the rectangle that we found, so with a little head scratching we can conclude that our estimate for height is a bit on the low side. Some of the extra area in the rectangle we drew that isn’t in region B will go into bumping the height up.

It turns out we can do even better, without touching trig or calculus, by finding another geometrical approximation to the area in region B, as you can see in the diagram below. We can go into details about solving it that way later (still no trig necessary!), but you should try it yourself, and see if you get any closer to h=4.04 inches!

For our Chopped Square puzzle, we attempted to find the side length of a square that was chopped up into 6 congruent rectangles.

Given a perimeter of 112 units for each of the six rectangles, what’s the length of each side of the square? Algebra steps in to help us out here.

Let’s first gather what we know.

Here is that square again:

First, the length of the small sides of each rectangle are equal to 1/6 of the length of the each side of the square. We know this because the rectangles are congruent, which means they have equal sides. If the shorter sides of the rectangles are all equal, and six of them make up the side of the square, then each small rectangle side is 1/6 the length of the square’s side. Let’s assign this 1/6 length a variable: x.

Second, the larger side of each rectangle is equal to the side of the square. This is immediately apparent from looking at the picture. Since we established above that the length of the square’s side is 6 times the length of the shorter rectangle side, we know that the larger rectangle side is also 6 times the length of the shorter one. Thus, the larger side can be expressed in terms of the shorter side: 6x.

The perimeter of a rectangle is length + length + width + width, and now we can fill those in with the terms we arrived at:

6x + 6x + x + x

We set that equal to 112, since that was the given perimeter of the rectangle:

6x + 6x + x + x = 112

Now, add together the like terms, which, in this case, is everything on the left side!

14x = 112

Divide 112 by 14, and we find out that x is equal to 8.

Remember that x is the length of the shorter rectangle side, and we set out to find the length of the square side, or the larger rectangle side. Since the larger rectangle side is 6x, all we have to do to get the answer is plug in 8 for x. That product, and therefore our answer, is 6 · 8 = 48!

We explore a similar problem in this tutorial here, to help you get the hang of it.

There are  other ways to chop up the square and find out more dimensions from what little information we were initially given.

How did you solve the problem? Let us know in the comments!

For today’s puzzle, we’ll have a nice blend of geometry and algebra, the visual and the abstract.  Geometry is easily applicable to the real world, whether you’re playing billiards or building a skyscraper.  And it’s often the case that a bit of proficiency in algebra is required to solve a geometry problem.  Hence, algebra also affects our daily lives.  Hopefully, this puzzle (and ones to follow) will give you a boost towards mastery of these two math disciplines.

The square shown below is cut into six congruent rectangles.  If two shapes are congruent, that means that they have exactly same side lengths and interior angles.  In other words, they are exactly the same size and shape!  So, given that the perimeter of each of these six rectangles is 112 units, what is the side length of the larger square?

If you need to brush up on the perimeter of a rectangle, just check out our tutorial here. Good luck and have fun!

Our fruit vendor sets his prices according to the number of vowels and consonants in the name of the fruit. For each vowel, the price of the fruit is raised $0.07, and for each consonant, the price is raised $0.03. In the form of an algebraic expression, his method looks like this:

Price = $0.07v + $0.03c

The number of vowels in any given fruit name is represented by the variable v, and the number of consonants is represented by the variable c. It’s an unusual method for determining prices, surely, but maybe the vendor is a math nerd and can’t help it.

Since even the best at math make mistakes, let’s check his numbers. A banana has three vowels and three consonants. Let’s plug those numbers into the equation:

Price = $0.07(3) + $0.03(3)

Multiply first, and then add, according to order of operations:

Price = $0.21 + $0.09
Price = $0.30

He got his price right! While we’re at it, let’s offer to write in his price for a pomegranate. What will it be? We have to solve our equation with v equal to 5 and c equal to 6, corresponding with the number of vowels and consonants in the word “pomegranate:”

Price = $0.07(5) + $0.03(6)
Price = $0.35 + $0.18
Price = $0.53

So each pomegranate costs $0.53. For your effort, the vendor gives you one for free. Enjoy!

It’s a beautiful Sunday morning and you’re headed to the local farmer’s market. When there, you spot a vendor selling fruit… but at quite odd prices. You walk up to inspect the price chart closely. Here’s what you see:

Banana: $0.30 each
Apple: $0.23 each
Pear: $0.20 each
Plum: $0.16 each

What method is the vendor using for determining prices, and what will be his price for a pomegranate, which he has yet to add to his sign?

It was noted that this is first a puzzle and secondly an algebra problem, since algebra is not needed to find the answer! For all of the size and variables this expression has, it actually simplifies to zero! This is because the sequence includes the binomial term (x – x), which equals zero. Anything multiplied by zero equals zero, and all of the terms in this sequence are being multiplied together.

However, you can still arrive at this conclusion by using algebra and FOILing all of the terms out. To show this, I will use just a portion of the original sequence: (x – v) · (x – w) · (x – x).

Using the FOIL method on the first two terms, (x – v) · (x – w), gives us x² – xw – xv + vw. After that, we have the expression (x² – xw – xv + vw) · (x – x). How do we FOIL those two? Well, FOIL is a handy acronym for remembering to multiply the individual terms of one binomial by the individual terms of the other binomial. This general idea extends to trinomials and polynomials and is called the distributive property. We explore this idea further in this tutorial and others. So let’s make it work here…

Multiplying the first four terms by the first x in (x · x), we get:

x³ – (x²)w – (x²)v + vwx

Then we multiply the four terms by the second, negative x, and have this result:

-x³ + (x²)w + (x²)v – vwx

As you can see, the negative x flipped the signs of the first result to give us our second result. When we add these two sets of four terms together, each term cancels with its counterpart of the opposite sign:

x³ – x³ = 0
-(x²)w + (x²)w = 0
-(x²)v + (x²)v = 0
vwx – vwx = 0

So, we have 0 + 0 + 0 + 0. That’s an emphatic zero! Armed with this answer, we can easily see what will happen when we include the rest of the original sequence. Even if all of the terms leading up to (x – x) are distributed, multiplying what we get by (x – x) will give us a bunch of numbers that add up to zero. Any multiplication thereafter we know will also give us zero. From (x – a) to (x – z), in multiplication, (x – x) trumps all!

For today’s puzzle, we have the following multiplication sequence:

(x – a) · (x – b) · (x – c) · (x – d) … and it continues on in this manner through the alphabet, ending with … (x – y) · (x – z). What does this expression simplify to?

The individual terms being multiplied are known as binomials, which you can learn more about here. We also have a tutorial about the FOIL method for multiplying them.

Keep in mind, and this is a hint, that this is first and foremost a puzzle and not necessarily an algebra problem. Good luck!

Thanks to Kevin Stone at Brainbashers.com for the puzzle idea.

Okay, so we let this one slide a bit. Sorry about that. This puzzle was originally published quite a while back, and we never got around to posting the answers. Time to correct our mistake.

First, a reminder of the rules…

Here are the rules. You must find an equation with the given numbers; inequalities would be too easy.  The numbers have to be used in the order in which they appear, as they appear.  No zeroes can be added.  You can use any mathematical operations, signs, or parentheses you want, as long as no numbers are added (such as from a cube root), but a square root, i.e., a radical sign, is acceptable.

OK, here we go:

1. Your friend calls and tells you they will arrive in half an hour.  You hang up and look at your watch: 3:32.  What equation can you use to remember the time at which they are expected to arrive, 4:02?

2. You realize that between the time your friend above said they will arrive in 30 minutes and the time you hung up, one minute of conversation passed.  It turns out they plan to be here at 4:01.  What equation can be made out of that?

3. Pizza is for lunch!  Your mom stuck it in the oven and asked you to take it out 19 minutes from now, at which time it will be 12:39.  Can you think of an equation for it?

4. You have a relative whose birthday you can never remember.  Perhaps an equation for it will help.  The date is 7/22/81.  Are you going to miss their next birthday?  Try and find two answers.

And here are the answers…

1. 4:02 can be written as √4 – 0 = 2

2. 4:01 can be written as 4^0 = 1. If you’re confused about 4 being raised to the power of zero, or anything related to exponents, see this Virtual Nerd tutorial.

3. 12:39 can be written as (1 + 2) · 3 = 9.

4. 7/22/81 can be written as 7 + 2 – 2 – 8 = -1 and as
(7 + 2)^2 = 81

Can you think of other solutions?

These are just examples; the real challenge is doing this whenever you want or need to, with dates and times for which it may be impossible. Is your birthday “equatable?” How about that of your friends or family members? Feel free to show off in the comments!

If you enjoy playing games with math, the way we do, you’ll see numerical relationships everywhere.  For example, when turning to page 64 in a book, you might note the number 64 as a perfect square, a perfect cube, and a number which has many factors.  A fun way to project math on the world around you is with the numbers of a date or time.

Often, when you see a date, such as 7/14/83, you can not only see the individual numbers and simple relationships, like how 7 is prime or 14 is 7 • 2, but that all of the numbers can be related to each other via an equation.

For example, 7/14/83 can be seen as 7 • 1 + 4 - 8 = 3.  It’s actually a great way of remembering dates, not to mention a way of brushing up on your arithmetic and keeping your mind sharp.

If you have a friend or relative born on 3/12/05, you can remember it every time by an equation that it forms: 3 • 1 + 2 = 05.

If any of this math seems a little rough, keep in mind the order of operations and how to add and subtract positive and negative numbers.

This can also be done with clock times, such as 12:42.  It can be turned into 1 • 2 - 4 = -2.

When it’s 3:29, you have 3² = 9.

If you need to remember an appointment, or the time dinner needs to come out of the oven, this is a great way to keep that time in memory.

Let’s give it a shot, shall we?

Here are the rules. You must find an equation with the given numbers; inequalities would be too easy.  The numbers have to be used in the order in which they appear, as they appear.  No zeroes can be added.  You can use any mathematical operations, signs, or parentheses you want, as long as no numbers are added (such as from a cube root), but a square root, i.e., a radical sign, is acceptable.

Here we go:

1. Your friend calls and tells you they will arrive in half an hour.  You hang up and look at your watch: 3:32.  What equation can you use to remember the time at which they are expected to arrive, 4:02?

2. You realize that between the time your friend above said they will arrive in 30 minutes and the time you hung up, one minute of conversation had passed.  It turns out they plan to be here at 4:01.  What equation can be made out of that?

3. Pizza is for lunch!  Your mom stuck it in the oven and asked you to take it out 19 minutes from now, at which time it will be 12:39.  Can you think of an equation for it?

4. You have a relative whose birthday you can never remember.  Perhaps an equation for it will help.  The date is 7/22/81.  Are you going to miss their next birthday?  Try and find two answers.

Here’s the explanation for the puzzle of the mysterious triangles…

This puzzle is made possible by the fact that the large, composite triangles are not actually triangles at all!  They are quadrilaterals (provided you fill in the 2 by 2 space in the lower one).  Any right triangle must have a base, height, and hypotenuse.  But the hypotenuse on these “triangles” is really made up of two different line segments.

The smaller blue and green triangles must have the same slope in order to form a continuous hypotenuse for the larger triangles.  Yet, when you calculate the slope for each, you will find that they are different.  The blue triangle has a slope (or “rise over run”) of ⁴/₁₀, which reduces to ²/₅.  The green triangle has a slope of ⁶/₁₆, which reduces to ³/₈.  ²/₅ is greater than ³/₈ and, thus, these slopes are not equal! If you forgot how to compare fractions we’ve got a great tutorial here.  If you look closely, you can see how the arrangement of the individual shapes affects the combined shapes, by either making the “hypotenuse” bulge out or cave in, and that’s where our extra space comes from: