Here’s the third match in Round 1 of The Big Internet Math-Off. Today, we’re pitting Matt Enlow against Sam Kay.

Take a look at both pitches, vote for the bit of maths that made you do the loudest “Aha!”, and if *you* know any more cool facts about either of the topics presented here, please write a comment below!

## Matt Enlow – Triangle of Medians

This excursion begins with a simple wondering I had several years ago: **If you take the three medians of a triangle (the segments joining each vertex to the midpoint of the opposite side), can you always make a new triangle having those medians as its three sides?**

(Those who are so inclined might pause here and form an opinion as to the answer before proceeding.)

The answer, it turns out, is “Yes!”

Now, I hope that you don’t take the above animation as anything resembling proof of my claim. Who knows what kind of trickery I might have employed? Maybe the specific triangle I chose to use is special in some way, and it doesn’t work for *all* triangles. Maybe I altered the lengths or angles of the medians as they slid across the screen. Don’t let me off the hook!

So how could we show that this is true in *all* cases?

One way would be to use the formula for the lengths of the medians \(m_a\), \(m_b\), and \(m_c\) in terms of the original side lengths \(a\), \(b\), and \(c\):

\[m_a=\frac{\sqrt{2b^2+2c^2-a^2}}{2}\text{, etc.,}\]

then use those, along with the fact that \(a\), \(b\), and \(c\) satisfy the Triangle Inequality (the sum of any two side lengths is greater than the third), to show that \(m_a\), \(m_b\), and \(m_c\) also satisfy the Triangle Inequality.

… Like I said… that’s *one* way. But it’s not a particularly appealing one.

A much simpler way would be to think about the sides and medians as *vectors*.

Let’s let \(A\), \(B\), and \(C\) be the vertices of our triangle. The median from vertex \(A\) can be thought of as the vector \(\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})\):

Likewise, the median from \(B\) is the vector \(\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{BA})\), and the median from \(C\) is the vector \(\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{CB})\).

If we can show that the sum of these three median-vectors is zero, then that will suffice to show that they can be thought of as the three sides of a triangle. Fortunately, this is straightforward to do:

\[\begin{align} &\;\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)+ \frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{BA}\right)+ \frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right) \\[0.5em] =&\; \frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}+ \overrightarrow{BC}+\overrightarrow{BA}+ \overrightarrow{CA}+\overrightarrow{CB}\right) \\[0.5em] =&\; \frac{1}{2}\left(\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+ \left(\overrightarrow{BC}+\overrightarrow{CB}\right)+ \left(\overrightarrow{AC}+\overrightarrow{CA}\right)\right) \\[0.5em] =&\;\frac{1}{2}(0+0+0) \\[0.5em] =&\;0 \end{align}\]

A second, more visual proof requires taking six copies of your triangle (along with its medians), and making them into a hexagon. You can then clearly see a triangle whose sides are *twice* the lengths of the medians:

Scaling this triangle down by a factor of \(\frac{1}{2}\) gives us the desired “triangle of medians.”

I like to tell my students that the best questions lead not only to answers, but to more questions. So where can we go from here? Every triangle has a “median triangle.” **What else can be said about how these two triangles are related?**

This question is intentionally extremely open-ended; I’d love for you to let your own wondering take you down your own path.

The path I traveled led me to discover that **the area of the median triangle is \(\frac{3}{4}\) that of the original triangle.**

Again, if you are so inclined, take some time to think about why that might be true before proceeding.

As always, there are many ways one could go about showing that this is true. Below is an animation of my favorite, which makes use of the following facts:

- Joining the midpoints of the three sides divides the triangle into four congruent triangles, and
- drawing the three medians of the triangle divides the triangle into six regions of equal area.

I hope you enjoyed this excursion as much as I did! Here is a little preview of coming attractions, should I be blessed with the opportunity to share them with you in this competition:

**Quarter-finals:** How to use squares and the Golden Ratio to make a circle

**Semi-finals:** Using a calculator (and some trigonometry) to (metaphorically) pull a rabbit out of a hat

**Finals:** Just when you thought you’d seen all of the gems hidden in Pascal’s Triangle…

**Matt Enlow** teaches mathematics at the Dana Hall School in Wellesley, MA. You can follow him on X, BlueSky and Mathstodon.

## Sam Kay – Counting Down to Doomsday

I Know What Day Of The Week You Were Born On.

No, I’m not performing one of those scams where I state ‘Sunday’ and hope to impress one-seventh of the people reading this piece. My neat little party trick involves asking someone when they were born, and roughly 10 seconds later giving them which day of the week that was.

How on Earth is this possible?

There are two routes a keen mathematician would come up with; either I know that the Gregorian calendar has a 28-year cycle, and so remembering 28 years’ worth of days and dates is a doable task, or I take a modular arithmetic approach. The latter is clearly easier to do.

The Doomsday algorithm is a method, an algorithm, popularised by Group Theory legend John Conway that pinpoints exactly what day of the week any given day in recent history was or soon-to-be future would be. It begins with the day of the year that Conway deems Doomsday which will always be the last day of February, no matter if it’s a leap year or not. At the time of writing (2024) it is a Thursday.

The first part of the algorithm uses Doomsday to figure out the day of the week for any day in the same year. An astounding result emerges when considering the even months that follow: the 4th of April, 6th of June, 8th of August, 10th of October, and 12th of December all fall on Doomsday. That is, this year 4/4, 6/6, 8/8, 10/10, and 12/12 are all Thursdays. From here, for any desired date you can simply count up or down from Doomsday.

Suppose I am hosting a birthday party on the 10th of August and I need to let my friends know which night of the week to keep available. I know that the 8th is a Thursday, and so the 10th being two days later makes it a Saturday. They probably have the day off anyway.

I will give a quick note on modular arithmetic as it is the heart of this process. Since we are working with days of the week, there are only seven cycling objects we need to keep track of. That is, after counting seven days we arrive back at the day we started on, which essentially counts zero days. This process works under arithmetic modulo 7. It is convention to relabel the days in terms of numbers as such:

**0** – Sunday, **1** – Monday, **2** – Tuesday, **3** – Wednesday,

**4** – Thursday, **5** – Friday, **6** – Saturday.

For the eager Aperiodical enjoyers reading this on the day of release, today is Wednesday, now relabelled as 3. 10 days from now will be 3+10 ≡ 3+3 = 6 modulo 7, which is a Saturday. All this is saying is that 10 days from now will be the same day as 3 days from now, because 10 ≡ 3 modulo 7.

There are other key Doomsday dates to remember in the year; a common mnemonic is “working a 9-5 in a 7-11 store” because 9/5, 5/9, 7/11, and 11/7 all match with Doomsday too. If my bar steward tells me I need to work late for a fresher’s week event on the 27th of September, I want to know which day of the week I should catch up on sleep for. The mnemonic tells us the 5th of September this year is a Thursday, 4, and the 27th is 18 days away. 18 ≡ 4 modulo 7, and 4+4 ≡ 1 modulo 7. This leaves us with a Monday.

March’s neat trick is that all multiples of 7 are Doomsdays*. For January and February, it us up to each mathemagician to decide which reference dates to use. This is because they are different dependent on the leap year-ness. I like to note that in the years that are not leap years, February has all multiples of 7 being Doomsday and January follows the same as October in that the 10th is Doomsday. In the case of a leap year, February now has all multiples + 1 being Doomsday and January’s Doomsday now follows the same as April and July (4th/11th).

*yes, that means Pi Day is a Doomsday! No, Tau Day is not a Doomsday.

It might already be an impressive feat knowing what day of the week any day in the year is. But the whole trick of knowing a person’s birth-day requires knowing Doomsday for any year at hand. This requires a bit more finesse.

In order to do this, one notes that Doomsday 2000 was a Tuesday, and Doomsday 1900 was a Wednesday. The importance of this is best described with an example. Let’s say we wanted to work out Doomsday for fifteen years’ time, 2039. It is common knowledge that moving up one year in the calendar shifts the days of the week by one. We first take Doomsday 2000 and add the number of years. 39 more days than Tuesday is 2 + 39 ≡ 2 + 4 = 6 modulo 7, a Saturday.

But we mustn’t forget leap years! Leap years shift the week by two days each. One asks how many leap years are between 2000 and 2039; it’s a 39-year difference, and since leap years occur every four years that leaves us with 9 leap years slotted in. Saturday plus 9 days is 6 + 9 ≡ 6 + 2 ≡ 1 modulo 7, a Monday. Doomsday 2039 is a Monday.

Having to count how many years and leap years there are between 1900 and 2000 is tedious work though, and not eﬀicient. There is one trick I have come across that involves the multiples of 12. Another astounding fact is that the \(n\)th multiple of 12 years after a century has its Doomsday \(n\) days after the century’s Doomsday. For example, Doomsday 1948 is Wednesday + 4 days = Sunday. Doomsday 2072 is Tuesday + 6 days = Monday.

The final part is to put these two ideas together and test this out on your willing family and friends. Heck, try it out on your own birthday and ask your mum if it was right – she’s bound to remember. Take, for example, Alice’s birthdate on 10th June 1996: 10th June is 4 days more than Doomsday, and Doomsday 1996 is Wednesday + 8 days using the multiple of 12s trick. 3 + 8 ≡ 4, so 10th June 1996 is 4 + 4 ≡ 1, a Monday.

My closing thought will be a quick disclaimer. I did say this was a neat little party trick, so by all means perform this at parties. Just make sure that the party in question contains people that know the day of the week they were born on. Otherwise you end up looking like a bit of a creep.

**Sam Kay** is a maths student at Durham University where he hosts the Chalkboard Ultra podcast and spends too much time thinking about spinors. Outside of maths, Sam runs one of the university’s jazz bands. You can follow him on X, and Chalkboard Ultra on X and Instagram.

So, which bit of maths has tickled your fancy the most? Vote now!

**Match 3: Matt Enlow vs Sam Kay**

- Matt with the triangle of medians
- (51%, 178 Votes)
- Sam with Doomsday
- (49%, 173 Votes)

Total Voters: **351**

**This poll is closed.**

The poll closes at 08:00 BST tomorrow. Whoever wins the most votes will get the chance to tell us about more fun maths in the quarter-final.

Come back tomorrow for our fourth match in round 1, pitting Mats Vermeeren against Howie Hua, or check out the announcement post for your follow-along wall chart!

Eh ??

Love doomsday. Go for it Sam.

For Matt’s observation that the area of the “median triangle” = 3/4 of original triangle, here’s how I justified it:

In any given triangle, one median divides it into 2 triangles of equal area.

In the large version (with a hexagon made up of 6 copies of original triangle, and large median triangle scaled by lenghth factor of 2) the large median triangle is made up of 6 halves of the 6 original triangles.

Let area of original triangle = A; then the large hexagon = 6A

The large median triangle = 3A.

The non-scaled median triangle = 1/4 area of large median triangle [for similar areas, area ratio = length ratio^2]

Thus, non-scaled median triangle = 3/4 A

What a lovely problem to think about!

Really enjoyed both great pitches today… THANKS!

Awesome! I thought someone might go the hexagon route!

All the best

From Matt’s pictures, it is clear that, given a triangle, the triangle of medians of its triangle of medians is similar to the original one, scaled 3/4.