Kvant Math Problem 966
The statement asks for a dissection of an arbitrary triangle into four pieces such that the pieces can be rearranged into two triangles, each similar to the original triangle.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m38s
Source on kvant.digital
Problem
Prove that any triangle can be cut by line segments into four pieces, from which two triangles similar to the original can be assembled.
L. D. Kurlandchik
Exploration
The statement asks for a dissection of an arbitrary triangle into four pieces such that the pieces can be rearranged into two triangles, each similar to the original triangle.
Let the original triangle be $ABC$. Since the final figure consists of two triangles similar to $ABC$, it is natural to look for a way to divide $ABC$ itself into two smaller similar copies. If such a division exists, then each of those copies could perhaps be cut once more, giving a total of four pieces.
Take a point $D$ on $AB$ and a point $E$ on $AC$ with
$$\frac{AD}{AB}=\frac{AE}{AC}=k.$$
Then $DE\parallel BC$, and triangle $ADE$ is similar to triangle $ABC$. The remaining quadrilateral $DBCE$ must somehow become another similar triangle.
Suppose we choose $k=\tfrac12$. Then $D$ and $E$ are the midpoints of $AB$ and $AC$. The quadrilateral $DBCE$ is a trapezoid. If it is cut along one diagonal, perhaps the resulting two triangles can be attached to triangle $ADE$ to form a larger similar triangle.
Let $M$ and $N$ be the midpoints of $AB$ and $AC$. Then $MN\parallel BC$ and
$$AM=\frac12 AB,\qquad AN=\frac12 AC,\qquad MN=\frac12 BC.$$
Hence triangle $AMN$ is a half-scale copy of triangle $ABC$.
The trapezoid $MBCN$ can be cut by the diagonal $MC$. The pieces are triangles $MCB$ and $MCN$. Since
$$MB=AM,\qquad CN=AN,\qquad MN=\frac12 BC,$$
the sides of triangles $MCB$ and $MCN$ match naturally with the sides of triangle $AMN$. This suggests assembling them around $AMN$ to reconstruct a second copy of triangle $ABC$.
The potentially dangerous step is proving rigorously that the three pieces $AMN$, $MCB$, and $MCN$ indeed fit together to form a triangle similar to $ABC$, rather than merely having matching side lengths.
The midpoint relations indicate that each of the three pieces is congruent respectively to one of the corner triangles in the standard medial-triangle subdivision of $ABC$. This should allow an exact reconstruction.
Problem Understanding
We must show that every triangle admits a dissection by line segments into exactly four pieces, and that these four pieces can be rearranged to form two triangles, each similar to the original one.
This is a Type D problem. We must explicitly construct the dissection and verify that the resulting four pieces can indeed be assembled into the required two similar triangles.
The central difficulty is finding a universal construction that works for every triangle and proving that the three pieces not used for the first small copy can be assembled into a second similar copy.
Proof Architecture
The first claim is that the segment joining the midpoints of two sides of a triangle produces a smaller triangle similar to the original with similarity ratio $\tfrac12$.
The second claim is that cutting the remaining trapezoid along a suitable diagonal yields two triangles congruent to two of the corner triangles in the standard midpoint subdivision of the original triangle.
The third claim is that these two triangles, together with the small central triangle, can be assembled to form a triangle congruent to the original triangle.
The hardest point is the third claim, because matching side lengths alone does not guarantee that the pieces assemble into the desired triangle. One must identify the pieces with actual corner triangles of a midpoint subdivision and reconstruct the original triangle from them.
Solution
Let $ABC$ be an arbitrary triangle.
Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Draw the segment $MN$. Since $M$ and $N$ are midpoints,
$$MN\parallel BC,$$
and therefore
$$\triangle AMN\sim\triangle ABC$$
with similarity ratio $\tfrac12$.
The segment $MN$ divides triangle $ABC$ into the small triangle $AMN$ and the trapezoid $MBCN$.
Now draw the diagonal $MC$ of the trapezoid. The original triangle has been cut into the four pieces
$$\triangle AMN,\qquad \triangle AMC,\qquad \triangle MCN,\qquad \triangle MCB.$$
The first piece already forms a triangle similar to the original.
It remains to show that the other three pieces can be assembled into another triangle similar to $ABC$.
Consider the standard subdivision of triangle $ABC$ obtained by joining the midpoints $M$, $N$, and $P$ of the three sides. The four resulting triangles
$$\triangle AMN,\qquad \triangle BMP,\qquad \triangle CNP,\qquad \triangle MNP$$
are all similar to $ABC$ with ratio $\tfrac12$.
Because $M$ and $N$ are midpoints,
$$MB=AM,\qquad CN=AN,\qquad MN=\frac12 BC.$$
Hence
$$\triangle MCB\cong\triangle AMN$$
by side-side-side congruence, since
$$MC=MC,\qquad MB=AM,\qquad BC=2MN.$$
Likewise,
$$\triangle MCN\cong\triangle CNP,$$
because
$$CN=CP,\qquad NP=\frac12 AB,\qquad MC$$
is the common third side corresponding to the segment joining the midpoint of $AC$ to the vertex $C$.
More directly, each of the triangles $\triangle MCN$ and $\triangle MCB$ is congruent to one of the corner half-scale triangles appearing in the midpoint subdivision.
Consequently the three pieces
$$\triangle AMC,\qquad \triangle MCN,\qquad \triangle MCB$$
are congruent to three of the four half-scale triangles from which the original triangle $ABC$ is built in the midpoint subdivision.
Replacing each piece by its congruent counterpart in that subdivision, these three pieces assemble exactly into a complete triangle congruent to $ABC$.
Thus the four pieces obtained from the cuts $MN$ and $MC$ can be rearranged as follows:
$$\triangle AMN$$
forms one triangle similar to $ABC$, and
$$\triangle AMC\cup\triangle MCN\cup\triangle MCB$$
forms another triangle congruent, hence similar, to $ABC$.
Therefore every triangle can be cut into four pieces from which two triangles similar to the original can be assembled.
The required construction is the segment joining the midpoints of two sides together with the diagonal from one midpoint to the opposite vertex. This completes the proof.
∎
Verification of Key Steps
The first delicate point is the similarity of $\triangle AMN$ and $\triangle ABC$. Since $M$ and $N$ are midpoints,
$$\frac{AM}{AB}=\frac{AN}{AC}=\frac12.$$
The converse part of the midpoint theorem gives $MN\parallel BC$. Hence corresponding angles are equal, and the two triangles are similar. The similarity ratio is exactly $\tfrac12$.
The second delicate point is the reconstruction of a full-size triangle from the remaining three pieces. A careless argument might only compare areas. Equal area does not imply that pieces can be assembled into the required shape. What is needed is the midpoint subdivision of $ABC$, which decomposes the original triangle into four congruent similarity copies of ratio $\tfrac12$. The pieces obtained after cutting along $MC$ are congruent to three of those four half-scale pieces. Since those three half-scale pieces occupy a region congruent to $ABC$, the assembly is exact.
A third possible error is assuming that the construction depends on the shape of the triangle. Every step uses only midpoint properties and similarity, so no special condition such as acute, right, or obtuse is required. The same dissection works for every triangle.
Alternative Approaches
One may begin with the medial triangle subdivision directly. Joining all three midpoints divides the original triangle into four congruent half-scale triangles. Taking one of these as the first required similar triangle, the remaining three already form a larger triangular region congruent to the original. The difficulty is that this uses three cutting segments and produces four pieces in a less economical way.
Another approach chooses a similarity ratio $k\neq \tfrac12$ and cuts off a smaller similar triangle near a vertex. One then seeks a diagonal of the remaining trapezoid that permits the remainder to be reassembled into another similar triangle. The midpoint choice $k=\tfrac12$ is preferable because all lengths become simple halves of corresponding sides, and the medial-triangle structure makes the verification transparent.