Isosceles Triangulation

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worthawholebean

3017 posts

worthawholebean

#1 h May 1, 2008, 4:57 PM • 4 Y

Y by Davi-8191, icematrix2, Adventure10, Mango247

Let $\mathcal{P}$ be a convex polygon with $n$ sides, $n\ge3$ . Any set of $n - 3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\mathcal{P}$ into $n - 2$ triangles. If $\mathcal{P}$ is regular and there is a triangulation of $\mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $n$ .

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t0rajir0u

12167 posts

t0rajir0u

#2 h May 1, 2008, 5:28 PM • 5 Y

Y by thunderz28, icematrix2, Adventure10, Mango247, and 1 other user

Solution

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CatalystOfNostalgia

1479 posts

CatalystOfNostalgia

#3 h May 1, 2008, 7:00 PM • 2 Y

Y by icematrix2, Adventure10

My Outline

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the future

898 posts

the future

#4 h May 1, 2008, 7:45 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

Hmm weird is this the same answer?

I had all powers of 2 greater than 4 (4,8,16....)

and then the recursively defined sequence (3,5,9,...,) where $a_{m+1} = 2a_{m} - 1$ where a1=3 and its for n>=1

for base cases and it can be any of those odd bases times a integral power of 2

So I had like (3,4,5,6,8,9,10,12,16,17...)

it seems like the same answer but what if i totally worded it differently

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frodo

206 posts

frodo

#5 h May 1, 2008, 7:52 PM • 3 Y

Y by icematrix2, Adventure10, and 1 other user

I'm glad I could complete this one! A friend of mine at school also qualified for USAMO and said he could do this problem too.

I'm not certain how many points I'll get for it though, since I beat around the bush at first when I proved that $n=4k+3$ where integer $k>0$ doesn't fit the criteria. Later, I managed to prove that if $n$ is odd, all $2^a+1$ work and are the only solutions, where $a>0$ .

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the future

898 posts

the future

#6 h May 1, 2008, 8:00 PM • 3 Y

Y by icematrix2, Adventure10, and 1 other user

frodo wrote:

I'm glad I could complete this one! A friend of mine at school also qualified for USAMO and said he could do this problem too.

I'm not certain how many points I'll get for it though, since I beat around the bush at first when I proved that $n = 4k + 3$ where integer $k > 0$ doesn't fit the criteria. Later, I managed to prove that if $n$ is odd, all $2^a + 1$ work and are the only solutions, where $a > 0$ .

uhh i think 5 works...
which is 2^2+1

anyways, dang i seem to have gotten the right answer but i wrote it in such a bad format i hope they wont count off too much...

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Xantos C. Guin

2057 posts

Xantos C. Guin

#7 h May 1, 2008, 8:31 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

t0rajir0u wrote:

We claim that $P(n, n) = 1$ if and only if $n = 2^a(2^b + 1), a, b \ge 0$ .

minor technicality that I discovered during the test: $(a,b) = (0,0)$ yields $n = 2$ which violates $n \ge 3$ . But if we allow degenerate polygons, then $n = 2$ works.

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Math Geek

817 posts

Math Geek

#8 h May 1, 2008, 8:58 PM • 4 Y

Y by icematrix2, Adventure10, Mango247, and 1 other user

the future wrote:

Hmm weird is this the same answer?

I had all powers of 2 greater than 4 (4,8,16....)

and then the recursively defined sequence (3,5,9,...,) where $a_{m + 1} = 2a_{m} - 1$ where a1=3 and its for n>=1

for base cases and it can be any of those odd bases times a integral power of 2

So I had like (3,4,5,6,8,9,10,12,16,17...)

it seems like the same answer but what if i totally worded it differently

That's what I got. And I semi-proved that those were the only ones by proving that the only odd numbers that existed were one more than a power of 2.

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Ubemaya

1560 posts

Ubemaya

#9 h May 1, 2008, 9:04 PM • 4 Y

Y by icematrix2, Adventure10, Mango247, and 1 other user

Do you guys think that only a rigorous solution like t0rajir0u's with nice notation and stuff will get full points, or will a solution that just sorta states everything (e.g. if n is odd, n-1 must be a power of 2 because as you keep on dividing the small polygons into smaller ones, they must have an odd number of sides) also get 6 or 7 points?

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alt

261 posts

alt

#10 h May 1, 2008, 9:17 PM • 2 Y

Y by icematrix2, Adventure10

ARGG I got the right solution but I don't think my solution is rigorous enough

I loosely defined an trapezoid-like polygon and showed that each regular n-gon must be divided into these trapezoid-like polygons if all the sides were to be used in isosceles triangles.

In the end my solution was much like an essay

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timwu

973 posts

timwu

#11 h May 1, 2008, 9:32 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

I divided the conclusion into three parts:
1. $2^p$
2. $2^p + 2^q$
3. $2^p + 1$

It's easy to prove that they work (actually you only need to prove 1, since the other two are just its application), but it's hard to prove that others don't. I used a lot of weird notations and wordy paragraphs to prove the converse. All I can remember is this point (I incoporated the converse into the proof of part 3):

( $n$ is not in form of power-of-2 or sum of non-zero-power of 2) Labelling the vertices $0,1,...,n-1$ . WLOG let $0$ be connected to $x$ , if $x$ is not a power-of-2, then $x\geq \frac {n + 1}{2}$ , and $n - x$ is a power-of-2. Then I basically proved $n - x$ must be $1$ by some shaky argument and then the above "lemma". Yeah, I know I'm due for deduction on lack of rigor, but hopefully let it be light...

Math Geek wrote:

That's what I got. And I semi-proved that those were the only ones by proving that the only odd numbers that existed were one more than a power of 2.

How do you prove the even ones that don't work?

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rem

1434 posts

rem

#12 h May 1, 2008, 9:38 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

So this was a bit similar to IMO 2006 number 2. Pretty nice question.

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james4l

2172 posts

james4l

#13 h May 1, 2008, 10:40 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

t0rajir0u wrote:

Solution

Tiny flaw, though. If a = 0, and b = 0, we are left with a degenerate 2-sided polygon. Ah well, that's what I got too (except rectified the a,b both = 0)

Xantos C. Guin wrote:

t0rajir0u wrote:

We claim that $P(n, n) = 1$ if and only if $n = 2^a(2^b + 1), a, b \ge 0$ .

minor technicality that I discovered during the test: $(a,b) = (0,0)$ yields $n = 2$ which violates $n \ge 3$ . But if we allow degenerate polygons, then $n = 2$ works.

If I remember correctly, i think they stated that $n > 2$ in the beginning

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samath

664 posts

samath

#14 h May 1, 2008, 10:44 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

They stated that $n\ge 3$ , yeah.

Just to be safe, I wrote at the end that they couldn't both be equal to 0.

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the future

898 posts

the future

#15 h May 1, 2008, 10:48 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

timwu wrote:

How do you prove the even ones that don't work?

umm for even ones i said that if you could divide the even ones some amount of time by 2 such that it will be come odd (so if the even number is a*2^k, you can divide by 2^k) and if a is a solution then so is a*2^k but if a wasnt, the a*2^k isnt either...thats where ill probably lose points...

i manage to get the right answer and prove that it works for those ones but i can really prove the other ones dont work...

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minusonetwelth

225 posts

minusonetwelth

#56 h Aug 23, 2022, 10:33 AM

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The anser is all numbers of the form $(2^a)(2^b+1)$ . Let the polygon be defined by $A_1A_2\ldots A_n$ . We begin by showing that all powers of $2$ work.
Claim:
Even $n$ work if and only if $\frac{n}{2}$ works.

Proof of the claim.
From the definition of triangulation, it follows that each side $A_{2i}A_{2i+1}$ must be part of a triangle. If $n$ is even, it is easy to that this triangle must in fact be $\triangle A_{2i}{A_{2i+1}}A_{2i+2}$ , we will call it "small". So after drawing all the lines $A_{2i}A_{2i+2}$ in (we obtain $\frac{n}{2}$ triangles), we obtain a regular $\frac{n}{2}$ -gon. From here the result is immediate. $\square$
Now we look at numbers of the form $n2^kr$ , where $r$ is odd. Because of parity reasons, there exists a side $A_iA_{i+1}$ which cannot be part of a small triangle, say $A_1A_2$ . The only vertex which can make this triangle be isoceles is the point $A_{1/2(n+3)}$ . Now each of the polygons $A_1A_{n}A_{n-1}\ldots A_{1/2(n+3)}$ and $A_2A_3\ldots A_{1/2(n+3)}$ must be further disected. Using the small triangle argument it is easy to see that this can only happen when their number of sides is each a power of $2$ , which must be the same power of $2$ as they have equal sides. Hence, $n$ must be of the form $2^b+1$ as desired.

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huashiliao2020

1292 posts

huashiliao2020

#58 h Aug 5, 2023, 7:49 PM

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badly written sol

The answer is all numbers n of the form 2^a+2^b, a,b nonnegative. Notice 2n works if n works because you can take A_iA_{i+2} as a side which makes isosceles triangle A_iA_{i+1}A_{i+2}, which we'll classify as short triangles. It's evident that doing this for the entire polygon reduces each two segments into one, effectively halving the sides.

On the other hand, we show that a polygon with one long side and l shorter sides can be triangulated iff l=2^k (we would think of this by drawing any diagonal which motivates it by reading this proof backwards). From the 2n<->n footnote, we only need to consider odd # of sides polygon not of the form 2^k+1 and show they don't work (since this will effectively cover all evens as well), and also show 2^k+1 works. WLOG orient the longest side to be at the bottom. Since 2^k+1 is odd, for the longest side to form an isosceles triangle, it needs to connect to the top most vertex. But this reduces the problem to (2^k+1-1)/2=2^{k-1}, which we know works. On the other hand, if it's not 2^k+1 sides, then by connecting the same vertices we reduce to a problem that is the form $2^ij\forall j=2m+1$ , which by induction hypothesis we know cannot be triangulated. Hence all numbers of the form $2^a(2^b+1)$ work.

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trk08

614 posts

trk08

#59 h Aug 8, 2023, 3:21 AM

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horribly written

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Jndd

1417 posts

Jndd

#60 h Aug 23, 2023, 2:37 AM

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We claim that all the numbers of the form $2^m+2^n$ for nonnegative $m,n$ work.

Label the vertices of a regular n-gon from $0$ to $n-1$ . Notice that all edges must be part of some triangle, so edge $(k,k+1)$ , vertices taken mod $n$ , must be a part of one of the following three triangles
$\begin{align*} &(k,k+1) - (k+1, k+2) - (k, k+2)\\ &(k-1,k) - (k, k+1) - (k-1,k+1)\\ &(k,k+1) - (k+1, k+1 + \frac{n-1}{2}) - (k+1 + \frac{n-1}{2}, k)\\ \end{align*}$ When $n$ is even, we must use either option 1 or 2 repeatedly for all the edges, so we see that $n$ has a triangulation if and only if $n/2$ has a triangulation.

However, when $n$ is odd, we must use option 3 exactly once. WLOG, $(n-1,0)$ is the edge that uses option 3, so we can't have vertex $\frac{n-1}{2}$ be blocked by $(\frac{n-3}{2},\frac{n+1}{2})$ , so we need $2\mid \frac{n-1}{2}$ , giving $n=4c+1$ . Now, the polygon is split into two halves, and for each half, we must be able to connect consecutive edges, because we can't use option 3 more than once. From this, we can arrive at $n=2^t+1$ for some nonnegative $t$ .

Using the previous two cases, we can have $n=2^t(2^u+1)$ or $n=2^t$ , which is equivalent to what we claimed.

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peace09

5417 posts

peace09

#61 h Oct 8, 2023, 7:13 PM

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joshualiu315

2513 posts

joshualiu315

#62 h Oct 27, 2023, 11:22 PM

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The answer is $n=\boxed{2^a(2^b+1)}$ for nonnegative integers $a,b$ .

For the sake of definitions, begin with a regular $(2n+1)$ -gon with points $A_1, A_2, \dots, A_{2n+1}$ in that order. Define a \textit{small} triangle to be a triangle likewise to $A_1A_2A_3$ and define a \textit{large} triangle to be a triangle likewise to $A_1A_{n+1}A_{2n+1}$ .

Claim: If a $n$ -gon can be triangulated, so can a $2n$ -gon.

Proof: Notice that filling out the $2n$ -gon with small triangles will result in an $n$ -gon, which proves the claim. $\square$

Now, consider the aforementioned regular $(2n+1)$ -gon. Evidently, there must be at least one large triangle. Upon further inspection, we see that each of the large triangles contains the center and thus, there are only $1$ such triangle.

The large triangle splits the polygon into two $n+1$ sided polygons with $n$ equal sides. Then, it is clear that $n$ must be a power of two, so the only $k$ -gons with an odd number of sides must have $k=2^b+1$ for a nonnegative integer $b$ . Then, we can multiply this by any factor of $2$ , so our answer results.

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EpicBird08

1740 posts

EpicBird08

#63 h Dec 14, 2023, 7:45 PM

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Our answer is $n = 2^x (2^y + 1).$

We will first need the following:

Claim: $2n$ works if and only if $n$ works.
Proof: If $n$ works, then $2n$ works because we can connect two corners that are not connected but closest possible with an edge. This gives $n$ isosceles triangles (which we will call $1-1-2$ triangles) on the side. Now, if $2n$ works, we note that every edge is contained in some triangle. Suppose that two adjacent vertices of a $2n$ -gon are $A$ and $B,$ , where $A$ and $B$ are in the same triangle. There is no point diametrically opposite an edge on a $2n$ -gon, so to make an isosceles triangle, so the third point of this triangle $C$ must be adjacent to either $A$ or $B,$ creating a $1-1-2$ triangle. This repeats for every edge, giving a regular $n$ -gon in the middle. Since $2n$ works, that forces $n$ to work, as claimed.

Since $n = 4$ clearly works, all powers of $2$ work, at which point we can note that $2^k = 2^{k-1}(2^0 + 1).$ Therefore, we only need to consider odd $n.$

There are two parts to the problem from here: proving that odd numbers of the form $2^k + 1$ work, and proving that the rest don't.

Part 1: Proving that $2^k + 1$ work. We attach a construction for $k = 5;$ it is obvious how to generalize.

Part 2: Proving that the rest don't work. Suppose that we have adjacent vertices $A$ and $B$ in our $(2k + 1)$ -gon. Then they must be part of the same triangle. There are $2$ ways that this triangle is isosceles: the first is if $A$ and $B$ are in a $1-1-2$ triangle, and the second is if this triangle includes the point diametrically opposite segment $AB,$ which we will call $C.$ Call a triangle $ABC$ very long. Note that every $1-1-2$ triangle covers $2$ edges of the polygon, but there are $2k + 1$ edges. Therefore, there is exactly one very long triangle in such a triangulation.

This very long triangle splits the polygon into $2$ congruent regions, each with $k$ edges facing outward and $1$ edge being inside the polygon. Note that $k$ is even, otherwise we cannot include the outward edges inside our triangulation (triangles containing outward edges are forced to be $1-1-2$ ). Similarly, $\frac{k}{2}$ is even, and so is $\frac{k}{4},$ and so on. This forces $k$ to be a power of $2,$ so $2k + 1 = 2^{x} + 1.$

Therefore, the only odd $n$ that work are $n$ of the form $2^x + 1.$ As a result, the only $n$ that work are $n$ of the form $2^x (2^y + 1),$ as claimed at the beginning.

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shendrew7

792 posts

shendrew7

#64 h Dec 21, 2023, 10:50 PM

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The key intuition is to consider the "small sides" of each shape we are triangulating. We use casework based on the value of $n$ .

$n$ is even: Each side must be the leg of an isosceles triangle rather than the base. Consequently, we are forced to pair the sides off at every other vertex, creating an interior $\frac n2$ -gon. Repeating this, we see that an $n$ -gon is successfully triangulated $\iff$ an $\frac{n}{2^{v_2(n)}}$ -gon can be triangulated.

The only special case to consider if when $n$ is a power of two, but we note these always work as we can use this algorithm to obtain a square.
$n$ is odd: Let $n=2n_1+1$ . We pair off the sides again, but this time 1 is left out. We are forced to use this side as the base with the vertex being the opposite vertex of the $n$ -gon. Hence if $\frac{n_1}{2}<1$ , we're done, but otherwise it must be an integer.

Then let $n_1=2n_2$ . After the previous move, we are left to triangulate two congruent polygons with $n_2$ small sides and one long side. This long side is longer than any diagonal in this new polygon, so it must be used as the base. Hence, if $\frac{n_2}{2}<1$ , we're done, but otherwise it must be an integer.

We continue, concluding that we require $n=2^x+1$ for a positive integer $x$ .

Hence our solutions for $n$ are $\boxed{2^x, 2^x+1, 2^x\left(2^y+1\right) \forall x,y \in \mathbb{Z}^+}$ . $\blacksquare$

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Markas

105 posts

Markas

#65 h May 6, 2024, 8:43 AM

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Case 1: n is even

Claim: If n is even, it works if and only if $\frac{n}{2}$ works.

Proof: If $\frac{n}{2}$ works, we can easily draw isosceles triangles on the top of every side of the old polygon. Now we have to prove the other direction. This can happen if we just pair up the adjacent sides of the polygon and draw the edges between them. This is the only such triangulation since the center of the polygon lies on the perpendicular bisector of a side, but no vertex lies on this bisector.

Case 2: n is odd

Claim: If n is odd, n works if $n = 2^k + 1$ .

Proof: Let a triangle be called big if it passes through the center of the n-gon. There is exactly one big triangle. Now this triangle divides the n-gon into two equal polygons. When each equal polygon has $2^{p-1}+1$ vertices for some p, we can triangulate it. Then the outer perimeter of the equal polygons concurrent with the side of the n-gon must be triangulated into small isosceles triangles, and we must continue this process $\Rightarrow$ it is only possible when $\frac{n-1}{2}$ is a power of 2 $\Rightarrow$ n works if n is of the form $n = 2^m(2^l +1)$ .

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dolphinday

1318 posts

dolphinday

#66 h Jul 2, 2024, 7:15 PM

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The answer is all $n$ so that $n = 2^a + 2^b$ .
First we can show that a $n$ -polygon is dissectible into isosceles triangles if and only if a $2n$ -polygon is.
Note that in a $2n$ -polygon, the only isosceles triangle containing $V_iV_{i+1}$ is $\triangle V_{i}V_{i+1}V_{i+2}$ (or $\triangle V_{i}V_{i+1}V_{i-1}$ ).
So then this leaves us with the pairs of adjacent sides triangulated and a central $n$ -gon, which must be dissectible into isosceles triangles if the $2n$ -gon is. The reverse also holds true, clearly.
We can then show all $2^a + 1$ -gons are dissectible into isosceles triangles, with the attached construction.
Our construction also shows the impossibility of $n \neq 2^a + 1$ (which suffices as we can multiply by $2$ as much as we want).
This is because we are forced to (WLOG) draw in $\triangle PGH$ , and continuously group adjacent sides into triangles(thus dividing the number of sides by $2$ ) until we get a number of sides that is not a power of $2$ , from which we can't draw in any more triangles.

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Om245

163 posts

Om245

#67 h Nov 25, 2024, 2:12 AM

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Answer is $2^m(2^n+1)$ for any $m,n$ whole number.

Claim:
$2n$ work if and only if $n$ works

If $n$ work, we can just add one vertex at midpoint of arc of each side, and hence so $2n$ works. Suppose $2n$ works, we have triangle \empty{good} if both of it's equal side are side of polygon. Now we claim all side of polygon are covered by \empty{good} triangles.

Suppose some side of polygon is not coverd by \empty{good} triangle then as number of side are even, perpendicular bisector of side of polygon pass through midpoint of arc of opposite side (as we always get rectangle for even side polygon). Hence we can't have isosceles triangle with base side as polygon side. Therefore only way is all side of polygon are part of \empty{good} triangle. Which give us another polygon at side $n$ . Hence if $2n$ is possible then so $n$ .

Claim:
If an odd $n$ works, then $n = 2^k+1$

let $n=2m+1$ . All side of polygon can't be side of \empty{good} triangle, Assume side $V_1V_2$ is base for some isosceles triangle. Then note it has to be $\triangle V_1V_2V_{m+2}$ .

Main observation is now we can't connect any vertex from ${V_3,V_4, \cdots V_{m+1}}$ so set ${V_{m+3},\cdots V_{2m+1}}$
By same arrgument as used in first claim, we have to use all side of polygon in \empty{good} triangle. then we have triangle $\triangle V_2V_3V_4, \cdots \triangle V_{m}V_{m+1}V_{m+2}$ as \empty{good} triangle. Note for this we should have $m+2$ even, hence $m$ is even.
Now we observe that after this we have polygon with side $V_2V_4V_6\cdots V_{m+2}$ (for one side). And if we consider it's side, then by same logic we still have to make \empty{good} triangles only. Doing this many time, we will be always making \empty{good} triangle on exiting polygon made up. Hence we only increase $v_2(m)$ each time. We finally end up with $m = 2^k$ for some $k$ .
Hence $n=2^k+1$ for odd $n$ .

This give us only answer, are number of type $2^m(2^n+1)$ .

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eg4334

617 posts

eg4334

#68 h Jan 13, 2025, 4:33 AM

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The idea for this problem is super fun. The answer is all numbers of the form $\boxed{2^m(2^n+1)}$ for nonnegative integers $m, n$ . Call a number that can be triangulated "funny". Call an isosceles triangle consisting of two adjacent sides of a polygon an "edge" triangle. The problem is split into two main claims:

Claim: A even number $2n$ is funny if and only if $n$ is funny.
Proof: One direction is obvious by inscribing a $n$ -gon inside of the $2n$ -gon by making isosceles triangles out of adjacent sides. For the other direction, consider any side. Note that the triangle containing this side must have this small side as one of the two isosceles sides because $2n$ is even. Therefore we must have one corner cut out. Repeat this process around the $2n$ -gon, proving the claim. $\blacksquare$ .

Claim: A odd number $n$ is funny if and only if $n = 2^m+1$ for some integer $m$ .
Proof:
Direction 1: We wish to show that all $2^m+1$ are funny. Consider one side of this polygon and form an isosceles triangle with the diametrically opposite vertice. We then form these two hemispheres on the side (see the image in the beginning for a visual I don't feel like asying right now). We induct on these hemispheres. Note that the number of vertices in these hemispheres are also one more than a power of two. The base case is obvious, and to generalize to the next hemisphere (i.e. upping the power of two by one) by creating edge triangles on the sides of this new hemisphere. Thus by induction all $2^m+1$ work, because we can dissect each of these hemispheres into isosceles triangles.

Direction 2: We wish to show that all funny $n$ -gons with $n$ odd must satisfy $n=2^m+1$ . Consider the sides of this $n$ -gon: These sides are either part of an edge triangle or have the third vertice diametrically opposite. Note that because $n$ is odd we cannot have all edge triangles and must have at least one diameter triangle. But because it is a diameter and these cannot intersect, we have exactly one diameter triangle. Now consider the two hemispheres drawn. We wish to show all hemispheres must have a number of vertices one more than a power of two. Consider one side of these hemispheres adajcent to the long diameter drawn. Because we cannot have another diameter, this side must be part of an edge triangle. Continue this process around the hemisphere, forming another hemisphere. Therefore we have shown that if a $k$ vertice hemisphere is funny then the $\frac{k+1}{2}$ vertice hemisphere works as well. This process continues until we reach a triangle, i.e $3=2^1+1$ . Now we work backwards to get the previous one to be $2(2^1+1)-1 = 2^2+1$ . Now the result immediately follows by induction.

Both directions are proven so the claim is done. $\blacksquare$ .

This is enough to finish. All odd $n$ are of the stated form, and all even $n$ are also of the stated form so we are done.

This post has been edited 1 time. Last edited by eg4334, Jan 13, 2025, 4:33 AM

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Maximilian113

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Maximilian113

#69 h Jan 13, 2025, 4:46 AM

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Sketch:

Note that if $n$ was even, consider one of the edges of the polygon. Since it is part of an isosceles triangle, it cannot be the base since $n$ is even. Therefore, it must be paired with an adjacent edge, and using the same logic to the other edges reduces our problem to triangulating a regular $\frac{n}{2}$ -gon. Thus, if $n$ is even it works if and only if $\frac{n}{2}$ works, so we consider when $n$ is odd. Clearly it is impossible to pair all the edges, so there must be one edge that is the base of an isosceles triangle. Once we construct this triangle, clearly the legs must be the base of another triangle (since there is no way to fit another one of that side length in each of the remaining two quadrilaterals). This works if and only if these remaining quadrilaterals have an odd amount of vertices. Continuing this logic, the following must be true: Let $a_i$ be a sequence of numbers such that $a_1=n, a_i = \frac{a_{i-1}+1}{2}$ for $i \geq 2,$ then at some point $a_i$ converges to $1.$ It is easy to show that only $n=2^k+1$ for some $k$ works. Therefore, in general if it is possible to write $n=2^i+2^j$ for nonnegative integers $i, j$ then $n$ works (because all odds of the form $2^k+1$ work while $n=4$ also clearly works.)

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blueprimes

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blueprimes

#70 h Feb 24, 2025, 12:23 AM

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Solved during chem, really fun problem! I think this is a different solution.

We claim the answer is all $n$ expressible in the form $2^a + 2^b$ for nonnegative integers $(a, b) \ne (0, 0)$ . Label the vertices of the $n$ -gon $A_0, A_2, \dots, A_{n - 1}$ in clockwise order. All proceeding indices referring to vertices of the $n$ -gon are $\pmod{n}$ . Define a $k$ -bowl as a polygon $A_{i} A_{i + 1} \dots A_{i + k}$ where $k \le \left\lfloor \dfrac{n}{2} \right\rfloor$ .

$\textbf{Claim 1:}$ A $k$ -bowl $B_0 B_2 \dots B_k$ can be triangulated if and only if $k$ is a power of $2$ .

Proof. Consider a $k$ -bowl that can be fully triangulated. Clearly no other segment has the same length as $B_1 B_{k + 1}$ so $k$ must be even and segments $B_0 B_{k / 2}$ and $B_k B_{k / 2}$ are drawn. This yields a triangle and two $k / 2$ -bowls. Thus, a $k$ -bowl can be fully triangulated if and only if a $k / 2$ -bowl can be triangulated. This readily implies that the only $k$ that work are powers of two.

Returning to the original problem, first we show a construction for any $n = 2^a + 2^b$ . If $a = b$ , simply draw segment $A_0 A_{2^a}$ dissecting the $n$ -gon into two $2^a$ -bowls. By $\textbf{Claim 1}$ we can fully triangulate these polygons as wanted.

Now if $a \ne b$ , WLOG $a > b$ . Then we can draw triangle $A_0 A_{2^{a - 1}} A_{2^{a - 1} + 2^b}$ which dissects the $n$ -gon into a triangle and two $2^{a - 1}$ -bowls and one $2^b$ -bowl, each of which can be fully triangulated, so again this works.

Finally, we show necessity. In any $n$ -gon that can be fully triangulated, consider an arbitrary triangulation, and WLOG $A_0 A_p A_{p + q}$ where $p, q, r$ are positive integers is the isoceles triangle with $A_0 A_p = A_0 A_{p + q}$ that (not strictly) contains the center of the $n$ -gon. Then polygons $A_0 A_1 \dots A_p, A_p A_{p + 1} \dots A_{p + q}, A_{p + q} A_{p + q + 1} \dots A_0$ are bowls, and thus we require $p$ and $q$ to be powers of $2$ . But $n = 2p + q$ and hence is expressible in the form $2^a + 2^b$ . Our solution is complete.

This post has been edited 1 time. Last edited by blueprimes, Feb 24, 2025, 2:25 AM
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gladIasked

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gladIasked

#71 h Mar 28, 2025, 3:49 PM

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Outline: The answer is all $n=2^i+2^j$ for some nonnegative integers $i$ and $j$ . We can show that $n=2^i+1$ can be triangulated with some induction. It's also easy to show that if a $k$ -gon can be triangulated, so can a $2k$ -gon (draw a segment connecting every other vertex in the $2k$ -gon), so we're done.

We can also show that if $n$ cannot be triangulated (with isosceles triangles), $n/2$ cannot be triangulated either — this is easy to see by noting that every side must be paired with exactly one other side when $n$ is even. In particular, each side needs to be paired with one adjacent side, so we're back at the $n/2$ case.

Finally, when $n$ is an odd number not of the form $2^i+1$ , in any legal triangulation we must draw the ``large" isosceles triangle going directly across the middle of the $n$ -gon. We can show that the remaining two polygons cannot be triangulated with some induction.

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