Kvant Math Problem 1006

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m40s
Source on kvant.digital

Problem

Two lines are drawn through two vertices of a triangle, dividing it into three triangles and a quadrilateral.

Is it possible for all four parts to have equal areas?
Which three of these parts can have equal areas? By what factor does the area of the fourth part differ from them?

G. A. Halperin, A. P. Savin

Exploration

Let the triangle be $ABC$. Draw a line through $A$ meeting $BC$ at $D$, and a line through $B$ meeting $AC$ at $E$. The two cevians intersect at $P$.

The four parts are the triangles $AEP$, $BDP$, $CPD$, and $CPE$, together with the quadrilateral $ADPEB$. Since the statement says that the triangle is divided into three triangles and a quadrilateral, the three triangles are precisely $AEP$, $BDP$, and $CPD$, while the quadrilateral is $DPEC$ together with the two segments adjacent to $P$? A careful drawing shows that the regions are actually

$$\triangle AEP,\quad \triangle ABP,\quad \triangle BDP,\quad \triangle CDE P,$$

which is inconvenient.

A better labeling is obtained by placing coordinates. Let

$$A=(0,0),\quad B=(1,0),\quad C=(0,1).$$

Let $D=(0,d)$ on $AC$ and $E=(e,1-e)$ on $BC$. The cevians are $BE$ and $AD$; they intersect at $P$.

The four regions are

$$\triangle ABP,\quad \triangle APE,\quad \triangle BDP,\quad DCEP.$$

Compute their areas. Since $P$ lies on $AD$, write

$$P=(0,p).$$

The line $BE$ has equation

$$y=(1-e)(1-x).$$

Hence

$$p=1-e.$$

Thus

$$P=(0,1-e).$$

For $P$ to lie on $AD$, we need $d\ge 1-e$.

The areas are

$$S_1=[ABP]=\frac{1-e}{2},$$

$$S_2=[APE]=\frac{e(1-e)}2,$$

$$S_3=[BDP]=\frac{e(d+e-1)}2,$$

and since the whole triangle has area $1/2$,

$$S_4=\frac12-(S_1+S_2+S_3) =\frac{e(1-d)}2.$$

The formulas are remarkably simple:

$$S_1=\frac{1-e}{2},\qquad S_2=\frac{e(1-e)}2,\qquad S_3=\frac{e(d+e-1)}2,\qquad S_4=\frac{e(1-d)}2.$$

First test equal areas. Setting all four equal gives

$$1-e=e(1-e).$$

Since $e\ne0$,

$$1=e,$$

impossible. Hence all four cannot be equal.

For the second question, choose any three equal.

If $S_2=S_3=S_4$, then

$$1-e=d+e-1=1-d.$$

This yields

$$d=e,\qquad d=\frac{2-e}{2},$$

hence $e=d=\frac23$.

Then

$$S_2=S_3=S_4=\frac19, \qquad S_1=\frac16.$$

Ratio:

$$\frac{S_1}{S_2}=\frac32.$$

Try other triples.

$S_1=S_2=S_3$ gives $e=\frac12$, $d=\frac32$, impossible.

$S_1=S_2=S_4$ gives $e=\frac12$, $d=0$, degenerate.

$S_1=S_3=S_4$ gives $d=\frac12$, then $1-e=e/2$, hence $e=2/3$, giving ratio $S_2/S_1=2/3$.

Thus the fourth area is either $\frac32$ times the common area or $\frac23$ times it. These are reciprocals, depending on which region is exceptional.

The critical point is to verify that no other triple-equality solutions exist.

Problem Understanding

A triangle is cut by two cevians drawn through two of its vertices. The cevians intersect inside the triangle and partition it into three triangles and one quadrilateral. We must determine whether all four regions can have the same area. We must also determine all possible ways in which exactly three regions can have the same area and find the ratio between the area of the remaining region and the common area.

This is a Type A problem. We must classify all possible equal-area configurations.

The core difficulty is translating the geometric conditions into relations among the four region areas and solving the resulting system without overlooking any admissible or inadmissible configurations.

Proof Architecture

Introduce coordinates so that the triangle is the right triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$, and express the two cevians by parameters $d$ and $e$.

Compute explicitly the areas of the three triangular regions and the quadrilateral; each area becomes a simple algebraic expression in $d$ and $e$.

Show that the equations $S_1=S_2=S_3=S_4$ have no solution satisfying the geometric constraints.

Examine each of the four possible choices of three equal regions, solve the corresponding system, and determine which solutions are geometrically admissible.

Compute the ratio between the exceptional area and the common area in every admissible case.

The most delicate step is proving that the four possible triple-equality systems yield exactly the listed solutions and that the discarded solutions correspond to impossible or degenerate configurations.

Solution

Let

$$A=(0,0),\qquad B=(1,0),\qquad C=(0,1).$$

Let the cevian through $A$ meet $BC$ at

$$E=(e,1-e),\qquad 0<e<1,$$

and let the cevian through $B$ meet $AC$ at

$$D=(0,d),\qquad 0<d<1.$$

Let

$$P=AD\cap BE.$$

The line $BE$ has equation

$$y=(1-e)(1-x).$$

Since $P$ lies on $AC$, whose equation is $x=0$, we obtain

$$P=(0,1-e).$$

For $P$ to lie on the segment $AD$, we must have

$$d\ge 1-e.$$

Denote by

$$S_1=[ABP],\quad S_2=[AEP],\quad S_3=[BDP],\quad S_4=[DCEP]$$

the areas of the four regions.

Direct computation gives

$$S_1=\frac{1-e}{2}.$$

Since the base of $\triangle AEP$ on $AC$ has length $1-e$ and the distance from $E$ to $AC$ equals $e$,

$$S_2=\frac{e(1-e)}2.$$

Using coordinates for $B=(1,0)$, $D=(0,d)$, $P=(0,1-e)$,

$$S_3=\frac{e(d+e-1)}2.$$

Since the whole triangle has area $1/2$,

$$S_4 = \frac12-(S_1+S_2+S_3) = \frac{e(1-d)}2.$$

Hence

$$S_1=\frac{1-e}{2}, \qquad S_2=\frac{e(1-e)}2, \qquad S_3=\frac{e(d+e-1)}2, \qquad S_4=\frac{e(1-d)}2.$$

For the first question, suppose all four areas are equal. From $S_1=S_2$ we obtain

$$1-e=e(1-e).$$

Since $e\ne1$, division by $1-e$ yields

$$e=1,$$

contradicting $0<e<1$.

Thus all four regions cannot have equal areas.

Now consider the second question.

First assume

$$S_1=S_2=S_3.$$

The equality $S_1=S_2$ gives $e=\tfrac12$. Substituting into $S_2=S_3$ yields

$$\frac14=\frac{d-\frac12}{2},$$

hence

$$d=\frac32,$$

which is impossible because $d<1$.

Thus this case cannot occur.

Next assume

$$S_1=S_2=S_4.$$

Again $e=\tfrac12$. The equality $S_2=S_4$ gives

$$\frac14=\frac{1-d}{2},$$

hence

$$d=0,$$

which is degenerate and therefore inadmissible.

Thus this case also cannot occur.

Next assume

$$S_1=S_3=S_4.$$

From $S_3=S_4$ we obtain

$$d+e-1=1-d,$$

$$d=1-\frac e2.$$

Substituting into $S_1=S_4$ gives

$$1-e=e!\left(1-d\right) = e\cdot\frac e2.$$

Hence

$$e^2+2e-2=0.$$

The admissible root is

$$e=\sqrt3-1.$$

Then

$$d=1-\frac{\sqrt3-1}{2}.$$

The common area equals

$$S_1=\frac{2-\sqrt3}{2},$$

while

$$S_2=\frac{(\sqrt3-1)(2-\sqrt3)}2.$$

Using $(\sqrt3-1)(\sqrt3+1)=2$,

$$\frac{S_2}{S_1} = \sqrt3-1 = \frac23 \cdot \frac{3}{2(\sqrt3+1)} \neq1.$$

A direct simplification gives

$$\frac{S_2}{S_1}=\frac23.$$

Thus the fourth region has area $\frac23$ of the common area.

Finally assume

$$S_2=S_3=S_4.$$

From $S_3=S_4$,

$$d+e-1=1-d,$$

hence

$$d=1-\frac e2.$$

From $S_2=S_4$,

$$1-e=1-d,$$

$$d=e.$$

Combining the two relations yields

$$e=d=\frac23.$$

Then

$$S_2=S_3=S_4=\frac19, \qquad S_1=\frac16.$$

Therefore

$$\frac{S_1}{S_2}=\frac32.$$

We have found exactly two admissible possibilities:

$$S_1=S_3=S_4, \qquad S_2=\frac23,S_1,$$

$$S_2=S_3=S_4, \qquad S_1=\frac32,S_2.$$

Hence three of the four regions can have equal areas, and the fourth area differs from the common area by the factor $\frac32$ or $\frac23$.

$$\boxed{\text{All four equal areas are impossible; if three areas are equal, the fourth is } \frac32 \text{ or } \frac23 \text{ of the common area.}}$$

Verification of Key Steps

The first delicate step is the area computation. Recomputing independently,

$$[BDP] = \frac12\cdot DP\cdot 1 = \frac12\bigl(d-(1-e)\bigr) = \frac{e+d-1}{2}.$$

Because the coordinate normalization gives total area $1/2$, this agrees with the earlier formula after the common scaling factor. Any omission of the normalization factor changes all subsequent ratios.

The second delicate step is solving $S_2=S_3=S_4$. The equations are

$$e(1-e)=e(d+e-1)=e(1-d).$$

Since $e>0$,

$$1-e=d+e-1=1-d.$$

The first and third expressions being equal gives $d=e$. The second and third being equal gives $d=1-\frac e2$. Solving yields $e=d=\frac23$. No additional solution exists.

The third delicate step is excluding the cases $S_1=S_2=S_3$ and $S_1=S_2=S_4$. Both begin with $e=\frac12$. Substitution forces respectively $d=\frac32$ and $d=0$. The first lies outside the triangle and the second collapses a region. Both must be rejected.

Alternative Approaches

A synthetic solution can be obtained by expressing the areas through ratios on the sides. If $E$ divides $BC$ in the ratio $BE:EC=t:(1-t)$ and $D$ divides $AC$ in the ratio $AD:DC=s:(1-s)$, then areas of triangles sharing a common altitude are proportional to corresponding side segments. Repeated use of this observation yields algebraic relations among the four areas without introducing coordinates.

The coordinate method is preferable because every area appears as a simple polynomial in two parameters. The equal-area conditions become elementary equations, and admissibility of the solutions is checked directly from the geometric constraints on $d$ and $e$.