Mind your decisions, I'm Presh Talwalkar
I'm thinking of two regular polygons
one polygon has m sides and D(m) diagonals and, the other has n sides and D(n) diagonals
If D(m) plus D(n) is equal to one hundred and twenty five,
solve for the value of m plus n
A version of this problem was asked in the 2018 MATHCOUNTS National Competition
Students have less than one minute to solve this problem without the use of a calculator.
I admit it took me about five minutes to solve. So, I think it's okay if you take longer to solve this problem.
I thank Patrick from Germany for the suggestion
Pause if you'd like to give this problem a try and when you're ready, keep watching to learn how to solve this problem.
I'm first going to present two formulas for the number of diagonals in a polygon with k sides
where k is greater than or equal to three
I'll then illustrate how to solve the problem quickly
One formula is an explicit formula
D(k) is equal to k multiplied by k minus 3 all over 2
This is useful if you want to evaluate a particular value like k is equal to one hundred
Another formula is known as the recursive formula.
This is D(k) is equal to D(k) minus one plus k minus two
This is useful for computing the next value if you know the previous value
This is useful if you want to create a table, which is what we need to do in this problem.
Before I get to that let me prove both of these formulas
I'll first prove the explicit formula by direct counting.
In a polygon with k sides, we have k vertices
How many diagonals can we draw from a particular vertex?
We'll notice that the adjacent vertices will form sides with this vertex
So these two sides will not be diagonals of this polygon.
We also cannot draw diagonals between the vertex and itself,
so these three vertices will be excluded in the count of drawing diagonals from this vertex.
But every other vertex will form a diagonal with this vertex.
Therefore, we have k minus 3 diagonals that we can draw from this vertex.
If we do this for all k vertices, we have k multiplied by k minus three
Now, we have to make one more adjustment.
Every single diagonal involves two different vertices.
To avoid double counting, we therefore have to divide this by two
And thus, we have k multiplied by k minus three all over two and that's one way to prove the explicit formula
Another way to prove it is by considering the complete graph
Consider k vertices as before
Now, let's connect every pair of vertices with an edge
In total we will have k choose two edges
because every pair of vertices will make an edge
Now, every edge is either a side or a diagonal of this polygon
but remember, there are k sides to this polygon.
If we subtract the second equation from the first we get a count of the number of diagonals
k choose two is equal to k multiplied by k minus one, over two
and when we subtract k,
we get k multiplied by k minus three all over two
and thus we've proven the formula in another way
So now, let me prove the recursive formula
Suppose we have k minus one vertices and we have a formula that counts the number of diagonals
How many more diagonals will we have, if we add one more vertex?
That is, if we have a k-sided polygon with k vertices.
Well, as before these three cases will not be diagonals of this new polygon
But every other connection will be a diagonal of the polygon.
So, we need to add on k minus 3 more diagonals.
Now, if we connect the adjacent vertices to make sides of this polygon,
let's examine what happens to this previous side of the previous polygon with k minus one vertices
We'll notice that this will actually be a diagonal of the new polygon.
So, we need to add one more to this formula
The result is that D(k) is equal to D(k) minus one plus, k minus three plus, one.
And that simplifies to be D(k) minus one plus, k minus two.
And that's the recursive formula we'll need to solve this problem quickly
So now, let's solve the problem quickly.
For fun, let's see how long it takes.
We'll first create a table for the number of diagonals
A triangle with three sides has zero diagonals; a quadrilateral has two diagonals
To get D(5), we take D(4) plus three.
This will be equal to five; then, D(6) is equal to D(5) plus four so, we get nine.
To get, seven we add five; to get eight, we add 6 and we quickly compute all of the values in this table
We want to exceed one hundred and twenty five diagonals.
Then, we look for a pair that has a sum of 125
This will be thirty five plus ninety corresponding to D(10) plus D(15)
Therefore, m plus n is equal to ten plus fifteen which is equal to twenty five.
And that's our answer! And we solved it in just under 40 seconds!
I think this is a really fun problem.
Thanks for making "Mind Your Decisions" one of the best channels on YouTube.
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