# Baseball’s Home Plate Is IMPOSSIBLE Mathematically

Hey, this is Presh Talwalkar Baseball is a game of traditions and rules. Rule 2.02 two of the MLB handbook carefully defines home plate by the following geometric construction. The home base is defined as a square where one side is 17 inches. Two corners are removed. The two adjacent sides measure 8 and 1/2 inches and the remaining two sides measure 12 inches and meet at a right angle. There’s only one problem with this very carefully defined construction It’s an impossible shape mathematically. This creates an interesting geometric little puzzle. If we were to say the dimensions of 17 inches and 8 and 1/2 inches as well as the following right angles are correct, what would be the correct measure of the remaining sides? That is, can you solve for the correct value of x to make home-plate geometrically possible? Can you figure it out? Give this problem a try and when you’re ready to keep watching the video for the solution. So how can we solve for the correct value of x? Will divide the shape into two different shapes. We’ll draw a line connecting the two sides of 8 and 1/2 inches. This creates an upper rectangle and a lower isosceles right triangle. Because the upper shape is a rectangle we know that this length will be 17 inches. Now we can focus on the isosceles right triangle. By the Pythagorean theorem x squared plus x squared is equal to 17 squared. This means 2x squared is equal to 17 squared, which means x squared is equal to 17 squared divided by 2. And so x is equal to 17 divided by the square root of 2, which is approximately 12.02, which is slightly larger than the 12 inches that’s defined in the handbook. If MLB wanted to be mathematically correct, it could say in the next iteration of its handbook that the two remaining sides should measure about 12 inches. Did you figure it out? Thanks for watching this video. Please subscribe to my channel I make videos on math. You can catch me on my blog Mind Your Decisions which you can follow on Facebook, Google+ and Patreon. You can catch me on social media @preshtalwalkar, and if you like this video please check out my books. There are links in the video description.

Been playing baseball since I was 5 (21 now) and never really thought about the math behind home plate! Always fun to learn more about the sport that has made up such a big part of my life.

I figured it out by the 45 45 90 triangle, each side is x, then hypo is x sqrt(2), which is equal to 17"

8.5×(2)^0.5=12.02

Took me all of 3 seconds to realize that the sides of the part cut away was 8.5, 8.5, and by identity, the length must be 8.5xsqrt(2)

Or you can go rectangle 16.973 x 8.5 rectangle and a Isosco

2 other possibilities 1.a rectangle 16.973 × 8.5 and a isoscele triangle with 12"× 12" × 16.973" neing square and width and length bieng of.027 of a inch small, which is like disecting a fine hair. Or 2 being out of squarre on each side where the two 12 inch sides of the Isosceles tringle meets the long end of rectangle bieng out of square .054" on each side.

This is really easy, I just:

x^2+x^2=17^2

2x^2=17^2

」(2x^2)=17

」2x=17

x=17/」2

You can calculate this also byusibg the triangle that was cut out. Its side is 8.5 inches so the hypotenusa is 8.5 * sqrt(2).

The difference is about half a millimeter so so the error is within measuring tolerances

If they use metric system like the rest of the world won't be approximate but it will all be exact to a mm

8.5/sin(45°)

Its simple. The answer is 12.02 so what is your problem. Most people would round down especially since .02 inches is only half a millimeter You try distinguish half a millimeter when cutting anything thicker than paper

You don't really need a calculator to show this. It's enough to show that sqrt(12^2+12^12)=12*sqrt(2), which cannot be simplified to 17. Assuming that the rules are interpreted correctly, but I don't even know what a "base" is.

I used the triangel that was cut off. sides need to be 8.5 ( 17 – 8.5 and 17 split in two for the corner ) Than a² + b² = c²

now i'm happy that for once I could work a question without extra help.

Yes I figured it out, fascinating.

Close enough.

17/sqrt(2)~=12,02…

Lengths given in anything except metres are only ever approximate anyway ….. Only measurements in SI units are accurate.

12.0208 inches. (A²+B²=C²)

A=8.5"

B (Which is half of 17") = 8.5"

Therefore C=12.02081528017131 (which I rounded down to 12.0208")

I sent this to the Major League baseball commissioner and he said he's going to have to cancel the rest of the season.

i hit 450 foot dingers into center field … .02 inches of homeplate is irrelevant

Mathematicians will never reach the home plate.

A question that I actually solved…

Neat

I Got It

2:45

aboutYou went way overboard with the complicated algebra.

Simple in your head math shows that forget the 12" and the home plate is perfectly matched 17/2=8.5 done.

It was done with inches not decimals.

I don’t see why you can’t just assume 17 inches in wrong, when you do, just using the Pythagorean theorem, you find out the hypotenuse should be 16.97 inches.

Got this one. Yay me.

12.02"

a² + b² = c²

a = 8½"

b = 8½"

All other measurements given had two sig figs. Therefore, the answer need only have two sig figs. 12 is fine as opposed to 12.02 because of that.

Well, 12.02 is 12 enough! =P

This is cake walk

Just unsubscribed for making me loose my time.

Oh poo…. a measurement of .02 in the 1800s? Get real. Do it the other way and the base should be 16.97. Nitpicky and silly.

Where does it say that the home plate must be planar? If it is cut from a hyperbola, this is definitely possible, and the curvature would likely not be noticeable.

I don't think that this puzzle is difficult enough to be on this channel.

Ok. Some things just dont need to be so precise. Play ball anyway.

Well, this shape is not possible in an Euclidien space …

With correct significant digits, x= 12 inches

Well, I'm just happy that I solved this one

I used to be really bad at math, but I'm trying very hard to become good

Presh: the sides can't be 12!

also Presh: they're

actually~12.02!.. cmon man. lol

Presh, at 0:34 you mention that MLB rule 2.02 mentions the 12” sides meet at a “right angle.” I believe you are mistaken.

I reviewed rule 2.02 you linked in the video description but do not find any reference to a “right angle” in the description of the home plate.

Can you provide the reference you actually used?

Were you assuming the 12” sides were supposed to meet in a right angle?

Cool

It is correct because it is limited by the smallest significant digits, IIRC. I came up with 16.97…, and that can only yield an accuracy of 17. Then there's tolerances and propagation of error. – things are getting foggy. Empirical testing will let one know if they chose correctly.

I solved it a different way. The remaining sides are 8.5 inches. Since both legs are equal in a right triangle you could then use the equation Xsqrt2. In this case 8.5sqrt2 is about 12.02

Improper sig figs

🤭

Mathematicians: I scoured the universe looking for inconsistencies. Guess what! Baseball home plates are 0.02 inches out!! That's a massive 0.166666666…% error!

Engineer: Meh. Thats close enough.

People watching baseball: What are those two pitch invaders doing with that ruler?

Much easier than most of the problems you give. 17/2 * sqrt(2) or ~ 12.02”

Yes

Actual value is 16.97 ~17

This was pretty easy

The MLB handbook says the dimensions are 17/8.5/12 and they meet at a point. The diagram in the appendix makes it appear that home plate is the bottom right triangle of the baseball diamond, but it doesn't put the "right angle" marker like it does for the other bases; only that the 1st base line and the 3rd base line form a right angle. Looking online, find the MLB groundskeeper blog where they state the point of home plate meets the imaginary lines from 1st and 3rd base. In reality, those lines are usually marked with a chalk line that is several inches wide, so there is quite a large tolerance for the "line".

You might be better to ask what angle is the point at home plate, you can use the cosine rule (note that interestingly 17^2 = 12^2 + 12^2 + 1 (I wouldn't be surprised if a mathematician was consulted when they were codifying the rules who pointed out that this shape very nearly worked… like fermat near misses on the simpsons). Anyway, the angle is actually 92.4.

Easy way a√2= 12 so a=6 six is not 8.5

There are so many way to solve this.

metric…Given that 12 is really anything greater than 11.5 and less than 12.5, it is mathmetically correct. Had it said 12.00 then your statement would be correct

The inventor of baseball was probably not good at math.

So he was like "who cares"

Or we could just make the rectangular sides 16.97 inches.

¿Que no?

Kin yew say "rounding error", boys and girls?

If the triangle's sides are taken as accurate, the angle cyphers to 90°11'56.2"

If the right angle is insisted upon, logarithmic error for the sides is 0.173% (1.73 micropartes)

Neither discrepancy is worth the math meany's tantrum. Namas té.

I used a different method, I considered x as the diagonal of a smaller square with the side of 8.5… Still resulted 12.02

Baseball is lame and so is this video

Did you know that pi isn't equal to 3.14159?

Excellent job baiting me for two one hundredths of an inch

I imagined the base was a square rotated to look like a diamond with two right triangles at the top then solved a^2+b^2=c^2 to get the same answer. Anyone else do the same thing?

My work: 8.5^2+(17/2)^2 =144.5

144.5^1/2

Can i use pythagoreom theorem, because I did and i got a completely different answer, around 54

2:49 x is wrong it should equal for the half of it

Ahhhh!! Bunch of nerd stuff…I'm out I'm heading back to conspiracy theories!!

You could have just as easily taken the size of the edges that were removed. Half of 17 is 8.5 so the missing length of the side is 8.5. 8.5^2+8.5^2 = 144.5.

√144.5 ≈12.02

Or if you divide the isoscele triangle in two by the 90 angle it can be sqrt 2 times 8.5 inch

I cracked in first attempt

Studying at VJTI MUMBAI

You should stick to cricket.

Better question, what kind of geometry would make this true?

Can be solved mentally by assuming hypotenuse of triangular part as Dia of a circle

12.0208" Given it starts with a square of 17", the two verticals are cut at 8.5", so the removed triangles are 8.5" right triangles. That gives the hypotenuse of (8.5^2 + 8.5^2)^0.5.

This is the most pointless video on the entire internet right now.

Pretty sure this doesn't matter because machines (or people) that make baseball plates are not accurate to the 1/50 inches anyway. .02 inches is less than than the thickness of a pencil lead.

The first problem that i can solve, lol

√8.5²+√8.5²=√x

Maybe the 0.2 inches extend outside the home plate

Let's just make the sides 8.5rt2 instead of 12, which essentially just be 12.02 inches instead.

Calculated a little bit differently but with same result. x = sqrt(8.5^2+8.5^2) ~ 12.02

That one was pretty easy

For anyone thinking it might be a problem rooted in turning metric measurements into imperial, it's not. The metric measurements state that the front edge is 43 cm and the triangle sides are 30 cm. However, The Pythagorean Theorem puts the triangular sides just over 30.4 cm.

Thankfully, the angles match up at least; the two angles connected to the triangular sides are 135° each.

I think this math is wrong as you said pythagorian theorem would be A squared + B squared = C squared. So 12 squared + 12 squared = Desired length squared

144+144=desired length squared. therefore 288 = desired length squared and since 17 squared = 289 the length should be just below 17 inches. Also the fact that the triangle is stated to be isoceles but when solved for is actually and equilateral triangle is kinda funny.

Do tell us how you propose to make a slab of rubber to 1/100 inch tolerance

x=(sin45)(17)

=12,02

Seems legit to me…lol

When your right 💪

Also you can do it using the hidden triangle that you cut out

I did it differently 8.5 x square root of 2

To make the shape geometrically possible, we could set the angle θ ≈ 90.20°, X ≈ 12.02", or side c ≈ 16.97". Rounding any of these results to the nearest whole number would result in the indicated dimensions which, while inaccurate in purely academic math terms, is perfectly acceptable in real world engineering terms as it falls within the indicated tolerances.

Sure, it's correct to say that 12 ≠ 12.02 but it is also correct to say that 12 ≈ 12.02 and that 12.02 is close enough to 12 that the difference is rather insignificant.

I calculated 12.031

The dimensions of the plate could be precisely correct with a truncated corner omitted which would have a length of .028"… Roughly.

That Like-to-Dislike ratio doesn't look good.

Even 12.02 isn't correct, you'd have to write the equation to be accurate, 12" is info enough for the game, this is a good little warm up problem tho

The use of the imperial system was the biggest error right from the start

People who don’t understand how significant digits are used should not be making videos about mathematics. Other videos from this person are not as bad so I guess they have just run out of material.

It would be 12.02 if the plate was 2 seperate shapes together, but its not, its one shape. the corners of the triangle do not extend completely as far as it would if it was a clean cut seperate shape. So when i measures out, you will lose approx .02 inches due to the slight radius that takes side of the isosceles into the side of the rectangle. Its practical math, not theoretical math, which in a way goes against the laws of mathematics, but is very real difference.

I’m sure when they manufacture the plates they just add the .02 inches to make it 90 degrees.

The guys that make them don't measure anything. They just mix, form, cut, box, and ship.

I have only watched 30 sec…. BOY I HATE INCHES/FEET/MILES….. Use the metric system, thanks

12.02081= x

Now solve the problem that is second base. For one, a hypotenuse of 127' 3-3/8" does not yield 90' to a side. More interestingly, home, first, and third all lie entirely within the 90' x 90' square (or diamond in baseball terms). Only 25% of second base lies inside the diamond (the center of 2nd is on the point of the diamond). Can we fix that? Anyone at MLB reading? First and third were moved over a century ago, time to move 2nd.