Splitting an Equilateral Triangle

We claim that N=2 is impossible and that every larger N is feasible.

First, we argue that N=2 is impossible: any subdivision into triangles must use piecewise-straight cuts. Since there are only two pieces there can be only one cut, and it cannot have any internal bends (as this would give a concave angle on one side), so it is a single straight line through the triangle. The two pieces therefore share one internal side, and to obtain a total of 6 sides the cut must pass through one vertex of the triangle and divide the opposite side into two (not necessarily equal) parts. We see easily now that if we avoid degeneracy then each piece is scalene: the untouched external side is strictly longer than the internal side, which is longer than the third.

To obtain N=3, join the centroid to each of the vertices of the triangle.

For larger values of N, first we note that any right triangle can be cut into two isosceles triangles by joining the right-angled vertex to the midpoint of the hypotenuse (which is the circumcenter of the right triangle). Thus any isosceles triangle can be divided into 4 isosceles triangles, by first dividing into 2 right triangles. We can therefore jump from any construction of N pieces to one with N+3 pieces.

Since N=1 is trivial and N=3 is possible, this covers two of three congruence classes mod 3. It remains only to give a construction for N=5: label the equilateral triangle ABC with centroid O. Let D be the midpoint of AC. Then line segments BD and OC divide the triangle into one isosceles (BCO) and two right triangles (ADB and CDO). The right triangles can be subdivided into isosceles triangles by joining D to the midpoints of AB and CO, giving 5 isosceles pieces.

An equilateral triangle cannot by dissected into 2 isosceles triangles, but if can by dissected into $n$ isosceles triangles for any $n \geq 3$ .

The proof consists of the following steps:

1) Observe that an equilateral triangle cannot be dissected into 2 isosceles triangles.

2) Observe that any isosceles triangle can be dissected into 3 isosceles triangles. In particular, an equilateral triangle can be 3-dissected in this way.

3) Conclude that, if any shape can be dissected into $n$ isosceles triangles, it can also be dissected into $n+2$ isosceles triangles.

4) Dissect an equilateral triangle into 4 isosceles triangles.

5) Conclude that an equilateral triangle can be dissected into $n$ isosceles triangles for any $n \geq 3$ but not for $n=2$ .

Let's elaborate on each of the steps:

1) A triangle can be dissected into two triangles only by slicing from a vertex through a point on the opposite edge. If ABC is an equilateral triangle and the cut goes from A to some point D on the side BC, then one of the angles at D must be at least 90 degrees; suppose this is the angle ADB. Then triangle ADB can only be isosceles if its angles at A and B are equal (the angle at D is too large); but the angle at B is 60 degrees and the angle at A is less than that, so ADB cannot be isosceles .

2) Given an isosceles triangle, it can be dissected into 3 smaller isosceles triangles using one of the following methods, depending on the character of the vertex angle: acute, right or obtuse. See this image for a quick visual explanation:

a) If the vertex angle is acute, drop a segment from the vertex along the vertical median until the left and right triangles become isosceles. This must happen by continuity: initially the segment is shorter than the inside edge, and eventually it's longer (since the median of an acute isosceles triangle is longer than half its base).

b) If the vertex angle is right, the above method fails by squishing the bottom triangle to degeneracy. This can easily be fixed, however, by dropping another altitude to slice one of the smaller right triangles.

c) Finally, if the vertex angle is obtuse, fire two rays from the vertex symmetrically down to the base, and swipe them until the left and right triangles become isosceles. Once again, this is guaranteed to happen by continuity.

3) This is easy: if any shape is dissected into $n$ isosceles triangles, pick one of the pieces and use the previous step to slice it into 3 smaller isosceles triangles. This increases the total number of parts by two (3 new triangles replace 1 original one).

4) Easy . Drop a median and then two more medians from its base.

5) Since an equilateral can be dissected into 3 and 4 isosceles triangles, and any such dissection can be refined by increasing the number of parts by 2, then any number of parts is feasible except 2. QED

(Note: Because I cannot submit diagrams through this form, I have tried to provide a clear description of the diagrams as follows.)

If there exists a dissection of equilateral $\triangle ABC$ into $n = m$ equilateral triangles, then we can also create a dissection into $n = m+3$ equilateral triangles. We do this by taking any existing triangle and subdividing it into 4 smaller equilateral triangles by cutting along the lines joining the midpoints of its sides.

Next, we observe the following construction creates a dissection of $\triangle ABC$ into $n = 2m$ equilateral triangles, for all $m \ge 2$ . First, locate points $A', B'$ on $AC, BC$ respectively, such that $\frac{AC}{AA'} = \frac{BC}{BB'} = m$ , and cut along $A'B'$ . Then divide $AB$ into $m$ equal parts, and $A'B'$ into $m-1$ equal parts, from which we may then dissect trapezoid $ABB'A'$ into $m+(m-1) = 2m-1$ congruent equilateral triangles of side length $AB/m$ . These, along with $\triangle A'B'C$ , gives the required dissection into $n = 2m$ equilateral triangles.

From both these constructions, it is easy to see that it is possible to find a dissection into equilateral (and hence isosceles) triangles for all positive integers $n$ except $n = 2, 3, 5$ .

For the cases $n = 3, 5$ , let $I$ be the incenter of $\triangle ABC$ and cut along $IA, IB, IC$ . This creates three congruent isosceles triangles with $IA = IB = IC$ . To make 5 triangles, locate $D, E$ on $AB$ such that $AD = DE = EB$ , and cut along $ID, IE$ . Then clearly $ID = IE = IB$ , so this dissects $\triangle AIB$ into two congruent isosceles triangles $\triangle ADI \cong \triangle BEI$ and one equilateral triangle $\triangle IDE$ .

Finally, we must show that no such dissection exists for $n = 2$ . Any such dissection may utilize only a single cut. Furthermore, this cut must pass through a vertex, since if not, it passes through two sides, creating four additional vertices for a total of 7; hence the result cannot yield two triangles, let alone two isosceles triangles. So suppose without loss of generality that the cut passes through $A$ and intersects at some point $P$ on $BC$ . Then $\angle ABP = 60^\circ$ , $\angle BAP < \angle BAC = 60^\circ$ , and $\angle APB = \angle ACB + \angle PAC > \angle ACB = 60^\circ$ . This cannot yield an isosceles triangle, so $N = \boxed{2}$ .

We first note that splitting into two isosceles triangles is impossible. The edges of the triangles either belong to the original triangle or are newly created (and so shared between the two triangles). Two triangular pieces of cheese can only share one edge without being stacked, so we can only draw one line segment: the shared edge. If the edge connects two of the original sides, we've made a trapezoid. If the edge connects a vertex and an edge, it is impossible to make two isosceles triangles, no matter the angle. Thus $N=2$ cannot be done.

Now, given an equilateral triangle, we consider two operations:

1. Connect the midpoints of the three sides into a triangle. This gives 4 equilateral triangles, 3 more than we start with.

2. Connect each vertex of the triangle to the centroid. This gives 3 isosceles (but not equilateral) triangles, 2 more than we start with.

By repeatedly applying the first operation, we can divide the cheese into $N$ equilateral triangles if $N=3k+1$ (1,4,7, etc.). If we apply the second operation after applying the first some number of times, we will add two additional triangles. For instance, we can turn four equilateral triangles into 6, 8, 10, or 12 triangles, depending on how many of the equilateral triangles we split into thirds. By applying the second operation once, we can achieve $N = 3k$ . By applying it twice, we can achieve $N=8,11,14,...$ . However, if we begin with 1 equilateral triangle, we cannot split it into thirds twice, and so we have not yet demonstrated that $N=5$ is possible.

Say ABC is our equilateral triangle. First pick $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ very close to $A$ ( $AD < \frac{1}{3} AB$ , say) with $\overline{DE}$ is parallel to $\overline{BC}$ . Triangle $ADE$ is equilateral, and will be one of our 5 pieces. Now consider a point $P$ somewhere on the altitude from $A$ to $\overline{BC}$ , below this line. No matter where it is placed, triangles $BPC$ and $DPC$ will be isosceles. If $P$ is on $\overline{DE}$ , we will have

$BP > BD > 2\cdot AD =2\cdot DP$ ,

because of our placement of $\overline{DE}$ . Similarly, if $P$ is on $\overline{BC}$ , we will have

$BP =\frac{1}{2} BC < \frac{2}{3}\frac{\sqrt 3}{2} BC < DP$ ,

considering the distance from $\overline{DP}$ to $\overline{BC}$ . Thus somewhere between these points we must have $BP=DP$ , and so triangle $BDP$ (and its mirror image) will be isosceles, and we will have split the original triangle 5 isosceles triangles.

Thus 2 is the only (and so maximal) $N$ that is impossible.

We will prove that, by induction that, for every $N \ge 6$ , we can divide the equilateral triangle into N equilateral triangle. Now, i will use $2$ times induction.

1. It's easy to check that, $N = 3, 4, 5, 6, 8$ satisfy the condition.

For $N = 3$ . Divide it into $3$ congruent isosceles triangle and define $D$ as circumcentre. See [url=http://www.geogebratube.org/student/m44155]here [/url]

For $N = 4$ , divide it into $4$ congruent equilateral triangle. See [url=http://www.geogebratube.org/student/m44153]here [/url]

For $N = 5$ . Lets, we make a equilateral triangle $CDE$ , and cyclic trapezoid $ABDE$ . And, assume that $F$ is the center of the circle. We obtain $5$ issosceles triangle. That is $\triangle CDE, \triangle EDF, \triangle DFB, \triangle AFB, \triangle AFE$ . See [url=http://www.geogebratube.org/student/m44151]here [/url]

For $N = 6$ . Divide into $6$ equilateral triangle. See [url=http://www.geogebratube.org/student/m44159]here [/url]

For $N = 8$ . Divide into $8$ equilateral triangle. See [url=http://www.geogebratube.org/student/m44161]here [/url]

1. It's easy to see that. If the triangle, can divide into $N$ equilateral triangle, then, we can divide the triangle into $N + 3$ equilateral triangle. Proof : Took one equilateral triangle. Then, since every equilateral triangle can divide into $4$ equilateral triangle. Then, divide this triangle. So, we obtain now $N + 3$ equilateral triangle

2. Since equilateral triangle can divide into $4, 6, 8$ equilateral triangle. Then, by induction and use the second step, for every $N \ge 6$ , we can obtain $N$ equilateral triangle.

3. So, by step $1$ and step $3$ , we obtain that for every $N \ge 3$ , we can divide an equilateral triangle into $N$ isosceles triangle.

4. We know that, a equilateral triangle cannot divide into $2$ isosceles triangle.

Proof : See equilateral triangle $ABC$ . WLOG, The only one possibility to make $2$ isosceles triangle is by define $D$ on $AB$ , such that we have two triangle $ACD$ and triangle $DCB$ . Since $CD, DA, BD < BC$ . Then, $AD = DC = BD$ . But, it's implies $CD$ is the altitude of the triangle. And, we know that $CD = DB \sqrt{3} > DB$ , which is contradiction with $CD = DB$ . So, we cannot make $2$ isosceles triangle

1. Then, he maximal number $N$ such that she cannot split the equilateral triangle block of cheese into $N$ non-degenerate isosceles triangles is $\boxed{2}$

I will assume all triangles in this proof are non-degenerate by default.

If you've already got an isosceles chunk of cheese there are two immediately available ways to split it into several smaller chunks. One is to cut from each vertex of the triangle to the circumcenter, which by definition has the same distance to all vertices, and thus each chunk will be isosceles with the two freshly cut sides being equal. This will make one chunk into three, with a net gain of $2$ chunks.

The other is to cut from midpoint to midpoint on all the sides. This will split the chunk into $4$ chunks, with a net gain of $3$ . However, while it may be intuitively clear, it's not mathematically obvious why the new triangles are isosceles. I will therefore digress for a paragraph while I justify this before I go back to helping Brilli.

Let's say we have a triangle $ABC$ . Let $A', B'$ and $C'$ be the midpoints opposite $A, B$ and $C$ , respectively. Then $AB$ is parallel with $A'B'$ , so the triangle $A'CB'$ has all the same angles as $ABC$ (note the ordering of the vertices) and they are therefore similar (scaled by a factor of $2$ ). The same goes for the other two triangles $AB'C'$ and $A'BC'$ . As a byproduct, we've also found that the sides of triangle $A'B'C'$ are all half as long as the corresponding sides in triangle $ABC$ , and it is therefore also similar to the big triangle. Thus all four small triangles are isosceles if the original triangle was.

Now back to the cheese. We start with one big chunk, and we've found that we can combine the two different cutting procedures to make either $2$ more chunks or $3$ more chunks, as much as we need to, and in any combination. It is not difficult to see that the only end result we cannot reach this way is $1$ more chunk, that is, two chunks total.

This begs the question, is there a way to cut the first cheese triangle $ABC$ into two isosceles cheese triangles? The answer is no. Assume for contradiction that it can be done, and note that if we are to cut one triangle into two triangles, we need the cut to go through one vertex (say $A$ ) and intersect the opposing side (let's say in the point $D$ ). The two resultant triangles will both share one angle and one side with the original triangle. But the angle they share is $60^\circ$ , and the only isosceles triangles to have a $60^\circ$ angle are the equilateral ones. That means that the triangles $ABD$ and $ACD$ are both equilateral. But they also share a side with the triangle $ABC$ . We conclude that both pieces are as big as the original triangle, and this is clearly absurd. Hence we cannot split one equilateral triangle into two isosceles triangles.

Therefore, the biggest integer $N$ such that the equilaterally triangular cheese cannot be split into $N$ isosceles triangles is $2$ .

By connecting the midpoints of an equilateral triangle, you split the triangle into four congruent equilateral triangles. Therefore, by continually splitting equilateral triangles in this way, you can split the triangle into 3k+1 parts (you start with one triangle, each division adds three). In addition, by bisecting each angle of an equilateral triangle, you can split the triangle into three isosceles triangles, thus adding two to the total. This allows you to take any of your divisions that yeild 3k+1 parts and make them yeild 3k+1+2 (or 3k+3) parts by bisecting each angle of one of the equilateral triangles.. Also, if you have already split the triangle into smaller equilateral triangles at least once, you can bisect the angles of two of the equilateral triangles. In other words, for any 3k+1 in which k is at least 1, you can achieve 3k+5. Therefore, we can split an equilateral triangle into any number of isosceles triangles that is congruent to 1 or 0 (mod 3) or any number congruent to 2 (mod 3) except for 2 and 5. However, we can split it into 5 parts by trisecting, for example, angle B of a triangle ABC, then splitting both angle A and angle C into two angles of value 20 and 40, with the 20 degree angle being on the side of angle A that includes line AB. Thus, we are only left with 2 as the unachievable number

Given an n-partitioned triangle, we can increase the number of partitions by 2. We can do this by taking a partition and dividing it into three isosceles triangles, using the circumcenter as the vertex where all three triangles meet. Thus we can divide the original equilateral triangle into N isosceles triangles for all odd N. We can also divide up the original equilateral triangle into four equilateral triangles by calling one of the sides the base and the other two sides legs and drawing a line parallel to the base through the midpoints of the two legs, and repeating this 2 more times by switching which side is the base. This means that now we can divide up the original equilateral triangle into N isosceles triangles for all N greater than or equal to 3. Now consider N=2. It is clear that this is impossible as the line segment that divides the equilateral triangle must be equal to each of the individual pieces of the side that it cuts. Hence N=2 is the maximal number that such a partition is impossible.

First, we show that it cannot be done with 2. If it can, by considering the sum of angles around the points, the extra vertex introduced must lie on the perimeter. It is easy to see that neither of these triangles can be isosceles.

For $N=3$ , it can be done via the circumcenter. For $N=4$ , it can be done via the midpoints. For $N=5$ , we first cut off a small equilateral triangle from one of the vertices, which leaves an isosceles trapezium. The trapezium is concyclic as the opposite angles sum to 180. Consider the vertices joined to the circumcenter of the trapezium. This gives us 4 isosceles triangles if the circumcenter lies within the trapezium. This is possible if the small equilateral triangle has length less than half the side of the original triangle.

Now, given any triangle, the altitude from the longest side lies within the triangle, and splits it into 2 right angled triangles. For each right angled triangle, taking the midpoint of the hypotenuse and joining it to the other vertex yields 2 isosceles triangles. Hence, we can go from 1 to 4. Thus, by nonstandard induction, we can get $N\geq 3$ .

Hence, the largest possible value is $N=2$ .

we will prove that for n>=2 we can always divide the equilateral triangle.

By connecting its centroid to its verticals, we have 3 triangles of 30-30-120.Let one of the new triangle be ABC with AB=AC. By drawing AH perpendicular to BC, and HD,HE perpendicular to AB,AC, we have 3 more isosceles triangles. Continue doing this proves that we can divide the original triangles into n=3k (k>=1) new isosceles triangles.

Similarly, by connecting 3 midpoints of 3 sides of the originial triangle and continue doing this, we can divide the original triangle into n=3k+1 (k>=0) new isosceles triangles.

Similarly one can prove n=3k+2(k>=1) also satisfies. Hence the maximum number of isosceles triangles that is unobtainable is 2.