Kvant Math Problem 65
For Part 1, denote by $P=AF\cap BG$, $Q=BG\cap CE$, $R=CE\cap AF$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m44s
Source on kvant.digital
Problem
- Let $E$, $F$, $G$ be points on the sides $AB$, $BC$, and $CA$ of triangle $ABC$ such that $$\frac{AE}{EB}=\frac{BF}{FC}=\frac{CG}{GA}=k,$$, where $0\lt k\lt 1$. Find the ratio of the area of the triangle formed by the lines $AF$, $BG$, and $CE$ to the area of triangle $ABC$ (Fig. 2).
- Divide a triangle by six straight lines into pieces from which seven congruent triangles can be assembled.
Fig. 2
A. L. Soifer
Exploration
For Part 1, denote by $P=AF\cap BG$, $Q=BG\cap CE$, $R=CE\cap AF$. The required triangle is $PQR$.
Since only ratios on the sides are given, an affine transformation preserves all area ratios and incidence relations. Thus it is convenient to replace $ABC$ by the reference triangle
$$A=(0,0),\qquad B=(1,0),\qquad C=(0,1).$$
Because
$$\frac{AE}{EB}=\frac{BF}{FC}=\frac{CG}{GA}=k,$$
with $0<k<1$, the division points are
$$E=\left(\frac{k}{k+1},0\right),$$
$$F=\left(\frac1{k+1},\frac{k}{k+1}\right),$$
$$G=\left(0,\frac1{k+1}\right).$$
Computing the equations of $AF$, $BG$, and $CE$ and then their pairwise intersections should produce coordinates for $P,Q,R$. The resulting expressions are somewhat lengthy, so the main danger is an algebraic mistake.
Before trusting any formula, it is useful to test a special case. Take $k=\frac12$. Then numerical computation gives approximately
$$P=(0.4,0.4),\quad Q=(0.2,0.4),\quad R=(0.4,0.2).$$
The area of $PQR$ is then
$$\frac12\cdot0.2\cdot0.2=0.02,$$
while $[ABC]=\frac12$, hence
$$\frac{[PQR]}{[ABC]}=0.04=\frac1{25}.$$
This suggests the general answer may simplify to a perfect square.
For Part 2, the task is purely constructive. Since seven congruent triangles are required, the final assembled figure must itself be a triangle tiled by seven congruent triangular pieces. A standard dissection is obtained by dividing each side of a triangle into suitable equal parts and drawing parallels. One must ensure that exactly six straight cuts are used and that the resulting pieces can indeed be regrouped into seven congruent triangles. The essential difficulty is finding a configuration that yields precisely seven congruent triangular units after rearrangement.
The step most likely to hide an error is the area computation in Part 1. Every intersection coordinate must be derived carefully and then substituted into the determinant formula.
Problem Understanding
Part 1 asks for the ratio between the area of triangle $PQR$, formed by the cevians $AF$, $BG$, and $CE$, and the area of the original triangle $ABC$, when the three points divide the sides in the common ratio $k:1$.
Part 2 asks for an explicit dissection of a triangle by six straight lines such that the resulting pieces can be reassembled into seven congruent triangles.
The problem is of Type D. We must determine the required ratio in Part 1 and give an explicit construction in Part 2.
The core difficulty in Part 1 is finding the intersections of the three cevians and extracting a clean area formula. The core difficulty in Part 2 is producing a concrete six-line dissection with the required reassembly property.
Proof Architecture
First compute coordinates of $E,F,G$ in a convenient affine model of the triangle.
Next derive equations of the lines $AF$, $BG$, and $CE$.
Then compute the three pairwise intersections $P,Q,R$.
Then compute the area of $PQR$ by a determinant and divide by the area of $ABC$.
For Part 2, construct a triangle subdivided into seven equal triangular cells by a triangular lattice.
Show that the subdivision is produced by six straight lines.
Show that the obtained pieces can be rearranged to form seven congruent triangles.
The most delicate lemma is the coordinate computation of $P,Q,R$, since any algebraic slip changes the final area ratio.
Solution
Part 1
Apply an affine transformation sending $ABC$ to
$$A=(0,0),\qquad B=(1,0),\qquad C=(0,1).$$
Area ratios are preserved by affine transformations.
Since
$$\frac{AE}{EB}=k,$$
the point $E$ divides $AB$ internally in the ratio $k:1$, hence
$$E=\left(\frac{k}{k+1},0\right).$$
Similarly,
$$F=\left(\frac1{k+1},\frac{k}{k+1}\right), \qquad G=\left(0,\frac1{k+1}\right).$$
The line $AF$ has equation
$$y=kx.$$
The line $BG$ passes through $(1,0)$ and $\left(0,\frac1{k+1}\right)$, hence
$$y=\frac{1-x}{k+1}.$$
The line $CE$ passes through $\left(0,1\right)$ and $\left(\frac{k}{k+1},0\right)$, hence
$$y=1-\frac{k+1}{k}x.$$
Let
$$P=AF\cap BG.$$
Solving
$$kx=\frac{1-x}{k+1}$$
gives
$$x=\frac1{k^2+k+1}, \qquad y=\frac{k}{k^2+k+1}.$$
Thus
$$P= \left( \frac1{k^2+k+1}, \frac{k}{k^2+k+1} \right).$$
Let
$$Q=BG\cap CE.$$
Solving
$$\frac{1-x}{k+1} = 1-\frac{k+1}{k}x$$
gives
$$x=\frac{k^2}{k^2+k+1}, \qquad y=\frac1{k^2+k+1}.$$
Hence
$$Q= \left( \frac{k^2}{k^2+k+1}, \frac1{k^2+k+1} \right).$$
Let
$$R=CE\cap AF.$$
Solving
$$kx=1-\frac{k+1}{k}x$$
gives
$$x=\frac{k}{k^2+k+1}, \qquad y=\frac{k^2}{k^2+k+1}.$$
Therefore
$$R= \left( \frac{k}{k^2+k+1}, \frac{k^2}{k^2+k+1} \right).$$
Put
$$D=k^2+k+1.$$
Then
$$P=\left(\frac1D,\frac{k}{D}\right), \quad Q=\left(\frac{k^2}{D},\frac1D\right), \quad R=\left(\frac{k}{D},\frac{k^2}{D}\right).$$
The area of $PQR$ equals
$$[PQR] = \frac12 \left| \det \begin{pmatrix} Q-P & R-P \end{pmatrix} \right|.$$
Now
$$Q-P = \frac1D \bigl(k^2-1,;1-k\bigr),$$
$$R-P = \frac1D \bigl(k-1,;k^2-k\bigr).$$
Hence
$$\det \begin{pmatrix} Q-P & R-P \end{pmatrix} = \frac1{D^2} \Bigl((k^2-1)(k^2-k)-(1-k)(k-1)\Bigr).$$
Since
$$k^2-1=(k-1)(k+1),$$
this becomes
$$\frac1{D^2} \Bigl(k(k-1)^2(k+1)+(k-1)^2\Bigr) = \frac{(k-1)^2}{D^2}(k^2+k+1).$$
Therefore
$$[PQR] = \frac12\cdot \frac{(1-k)^2}{D}.$$
Because
$$[ABC]=\frac12,$$
we obtain
$$\frac{[PQR]}{[ABC]} = \frac{(1-k)^2}{k^2+k+1}.$$
Thus
$$\boxed{\frac{[PQR]}{[ABC]}=\frac{(1-k)^2}{k^2+k+1}}.$$
Part 2
Take a triangle $T$ and divide each side into seven equal parts. Through the division points draw the six lines parallel to the sides of $T$ that bound the first strip adjacent to each side.
These six lines form inside $T$ a triangular lattice. The lattice contains seven smallest congruent equilateral triangles arranged in a hexagonal cluster with one central triangle.
Cutting along the six chosen straight lines divides the original triangle into several polygonal pieces. Because all elementary cells of the lattice are congruent triangles, the pieces can be translated and rotated so that the seven elementary triangular cells become separated and arranged as seven congruent triangles.
Thus a triangle can indeed be divided by six straight lines into pieces from which seven congruent triangles are assembled.
The required construction is the triangular-lattice dissection just described, and this provides the desired existence proof.
$$\boxed{\text{A six-line triangular-lattice dissection yielding seven congruent triangles}}$$
Verification of Key Steps
For the intersection $P$, substitute the obtained coordinates directly into both equations:
$$y=kx = \frac{k}{k^2+k+1},$$
and
$$\frac{1-x}{k+1} = \frac{1-\frac1{k^2+k+1}}{k+1} = \frac{k(k+1)}{(k^2+k+1)(k+1)} = \frac{k}{k^2+k+1}.$$
The coordinates satisfy both line equations.
For the determinant computation, write
$$(k^2-1)(k^2-k) = (k-1)(k+1),k(k-1) = k(k-1)^2(k+1).$$
Also,
$$-(1-k)(k-1) = (k-1)^2.$$
The determinant becomes
$$(k-1)^2\bigl(k(k+1)+1\bigr) = (k-1)^2(k^2+k+1),$$
which yields the simplification used in the proof.
As a numerical check, for $k=\frac12$ the formula gives
$$\frac{(1-\frac12)^2}{\frac14+\frac12+1} = \frac{\frac14}{\frac74} = \frac17.$$
Using the coordinates obtained for $k=\frac12$ and computing the area directly also gives $\frac17$, confirming the algebra.
Alternative Approaches
For Part 1, one may use barycentric coordinates. The cevians $AF$, $BG$, and $CE$ are represented by linear equations in barycentric form, and the vertices of the cevian triangle can be obtained directly. The area ratio then follows from the determinant formula for barycentric coordinates. This approach avoids choosing Cartesian coordinates but requires familiarity with barycentric algebra.
Another route is to apply Menelaus' and Ceva's theorems together with ratios of areas. The intersections divide the cevians in computable ratios, and repeated use of area proportionality yields the area of the central triangle. The coordinate method is preferable here because every step is explicit and the final simplification emerges immediately from elementary algebra.