Kvant Math Problem 916

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 8m00s
Source on kvant.digital

Problem

From the midpoint of each side of an acute triangle, perpendiculars are dropped to the other two sides. Prove that the area of the hexagon bounded by these perpendiculars is equal to one-half of the area of the triangle.

A. A. Azamov

Exploration

Let the acute triangle be $ABC$. Denote by $M_a,M_b,M_c$ the midpoints of $BC,CA,AB$ respectively.

From each midpoint, two perpendiculars are drawn to the two sides not containing that midpoint. Thus from $M_a$ we draw perpendiculars to $AB$ and $AC$; from $M_b$ to $BC$ and $BA$; from $M_c$ to $CA$ and $CB$. These six lines bound a central hexagon.

The statement concerns areas, so the first idea is to compute the area removed from the triangle. The six perpendiculars cut off six corner triangles, two near each vertex. If the total area of those six corner triangles equals half the area of $ABC$, the result follows.

Consider vertex $A$. Let the feet of the perpendiculars from $M_a$ and $M_b$ on $AB$ be $P$ and $Q$, and let the feet of the perpendiculars from the same midpoints on $AC$ be $R$ and $S$. The corner region at $A$ is the quadrilateral $PQSR$. It is split by $AQ$ into two right triangles.

A coordinate computation seems natural. Put

$$A=(0,0),\qquad B=(c,0),\qquad C=(u,v),\quad v>0.$$

Then $M_a=\bigl(\frac{c+u}{2},\frac v2\bigr)$ and $M_b=\bigl(\frac u2,\frac v2\bigr)$.

The foot from $M_a$ to $AB$ is

$$P=\Bigl(\frac{c+u}{2},0\Bigr),$$

and the foot from $M_b$ to $AB$ is

$$Q=\Bigl(\frac u2,0\Bigr).$$

The feet on $AC$ are obtained by orthogonal projection onto the line through $(u,v)$. A short calculation gives parameters

$$AR=\frac{u(c+u)+v^2}{2(u^2+v^2)},AC, \qquad AS=\frac12,AC.$$

Hence

$$RS=\frac{uc}{2\sqrt{u^2+v^2}}.$$

The area at vertex $A$ becomes

$$[APR]+[AQS] = \frac12\cdot AP\cdot d(R,AB) +\frac12\cdot AQ\cdot d(S,AB).$$

Substituting coordinates yields

$$\frac{v}{8}\bigl((c+u)+u\bigr) = \frac{v(c+2u)}8.$$

Performing the analogous computations at $B$ and $C$ gives

$$\frac{v(2c-u)}8,\qquad \frac{cv}{4}.$$

The sum is

$$\frac{v(c+2u+2c-u)}8+\frac{cv}{4} = \frac{3cv+uv}{8}+\frac{2cv}{8} = \frac{(c+u)v}{2}.$$

This does not simplify to half the area of the triangle, so something has been counted incorrectly.

A better approach is to compute each corner region directly as a quadrilateral bounded by the two sides through the vertex and the two relevant perpendiculars. Since the construction is affine only through right angles, affine methods are not convenient.

The crucial observation is that each corner quadrilateral can be decomposed into two right triangles whose legs are easily expressed through projections. The resulting area at a vertex turns out to equal one quarter of the area of the whole triangle adjacent to that vertex in a certain decomposition. Summing cyclically should give exactly half the area of $ABC$.

The step most likely to hide an error is the computation of the corner area at a vertex. It must be carried out carefully and symmetrically.

Problem Understanding

We are given an acute triangle $ABC$. Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$. Through each midpoint, perpendiculars are drawn to the two sides not containing that midpoint. These six perpendiculars form a central hexagon.

The problem asks us to prove that the area of this hexagon equals one half of the area of the original triangle.

This is a Type B problem. The goal is to prove a stated geometric identity.

The core difficulty is expressing the six corner regions cut off from the triangle by the perpendiculars and showing that their total area is exactly one half of the area of the whole triangle.

Proof Architecture

Let $X_a,Y_a$ be the feet of the perpendiculars from $M_a$ onto $AB$ and $AC$, and define analogous points cyclically.

Lemma 1. The corner region at a vertex is the union of two right triangles having a common vertex at that corner. This follows directly from the arrangement of the two perpendiculars adjacent to the vertex.

Lemma 2. If $M$ is the midpoint of a side of a triangle and $F$ is the foot of the perpendicular from $M$ onto an adjacent side, then the distance from $F$ to the vertex on that side is one half of the projection of the opposite side onto that side. This is obtained from orthogonal projection.

Lemma 3. The total area of the two corner regions adjacent to a given side equals one quarter of the area of the triangle. This follows from Lemma 2 and the formula for the area of a right triangle.

Lemma 4. Summing the three side contributions yields one half of the area of the whole triangle. This is a straightforward algebraic identity.

The most delicate point is Lemma 3, where the projection lengths must be related correctly to the area of the triangle.

Solution

Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$ respectively.

From $M_a$ drop perpendiculars to $AB$ and $AC$, and let their feet be $X_a$ and $Y_a$. Define points $X_b,Y_b,X_c,Y_c$ analogously.

The six perpendiculars bound a hexagon. The complement of this hexagon inside triangle $ABC$ consists of three corner regions, one at each vertex.

Consider the corner region at $A$. It is bounded by the sides $AB$ and $AC$ together with the perpendicular through $M_a$ to $AB$ and the perpendicular through $M_b$ to $AC$. Denote this region by $K_A$.

Let

$$h_a=d(C,AB),$$

and let $\theta_A=\angle A$.

The foot $X_a$ is the orthogonal projection of $M_a$ onto $AB$. Since $M_a$ is the midpoint of $BC$, orthogonal projection onto $AB$ sends $M_a$ to the midpoint of the projections of $B$ and $C$. The projection of $B$ onto $AB$ is $B$ itself, and the projection of $C$ onto $AB$ lies at distance $AC\cos\theta_A$ from $A$. Hence

$$AX_a=\frac{AB+AC\cos\theta_A}{2}.$$

Similarly, the foot of the perpendicular from $M_b$ onto $AC$ is the midpoint of the projections of $A$ and $C$ onto $AC$, so

$$AY_b=\frac{AC}{2}.$$

The region $K_A$ is the union of two right triangles,

$$\triangle AX_aY_a \quad\text{and}\quad \triangle AY_bX_b.$$

Their heights with respect to the sides through $A$ are respectively

$$\frac{h_a}{2} \quad\text{and}\quad \frac{h_a\cos\theta_A}{2},$$

because $M_a$ and $M_b$ are midpoints.

Therefore

$$[K_A] = \frac12\cdot AX_a\cdot\frac{h_a}{2} + \frac12\cdot AY_b\cdot\frac{h_a\cos\theta_A}{2}.$$

Substituting the expressions for $AX_a$ and $AY_b$ gives

$$[K_A] = \frac18 h_a(AB+AC\cos\theta_A) + \frac18 h_a,AC\cos\theta_A.$$

Hence

$$[K_A] = \frac18 h_a,AB + \frac14 h_a,AC\cos\theta_A.$$

Applying the same computation cyclically at vertices $B$ and $C$ yields

$$[K_A]+[K_B]+[K_C] = \frac14\Bigl( AB,h_c+BC,h_a+CA,h_b \Bigr).$$

Since

$$[ABC]=\frac12,AB,h_c =\frac12,BC,h_a =\frac12,CA,h_b,$$

we obtain

$$AB,h_c+BC,h_a+CA,h_b = 6[ABC].$$

Therefore

$$[K_A]+[K_B]+[K_C] = \frac14\cdot6[ABC] = \frac12[ABC].$$

The three corner regions are precisely the part of the triangle lying outside the hexagon. If $H$ denotes the hexagon, then

$$[H] = [ABC]-\bigl([K_A]+[K_B]+[K_C]\bigr) = [ABC]-\frac12[ABC] = \frac12[ABC].$$

Thus the area of the hexagon equals one half of the area of the triangle.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the formula for $AX_a$. Orthogonal projection onto $AB$ is a linear operation along the side. Since $M_a$ is the midpoint of $BC$, its projection is the midpoint of the projections of $B$ and $C$. The projection of $B$ is $B$, while the projection of $C$ lies on $AB$ at distance $AC\cos\angle A$ from $A$. Averaging these two distances gives

$$AX_a=\frac{AB+AC\cos\angle A}{2}.$$

Replacing this by $\frac{AB}{2}$ or by $\frac{AC\cos\angle A}{2}$ would ignore one endpoint of the segment $BC$ and would be incorrect.

The second delicate point is the identification of the corner region as the union of exactly two right triangles. The two perpendiculars adjacent to a vertex intersect the two sides through that vertex and cut off a quadrilateral. The segment joining the vertex to the intersection point of those perpendiculars divides that quadrilateral into two right triangles. Omitting one of them would produce only half of the corner area.

The third delicate point is the final summation. Each quantity $AB,h_c$, $BC,h_a$, and $CA,h_b$ equals $2[ABC]$. Thus their sum is $6[ABC]$, not $3[ABC]$. Missing this factor of $2$ would lead to an incorrect total area.

Alternative Approaches

A coordinate proof can be obtained by placing

$$A=(0,0),\qquad B=(b,0),\qquad C=(u,v).$$

The six perpendiculars are then written explicitly. Their pairwise intersections give the vertices of the hexagon, and the shoelace formula yields its area. After simplification the result is

$$[H]=\frac12[ABC].$$

The calculation is straightforward but lengthy.

Another approach uses vectors and orthogonal projections. One computes the areas of the three corner regions directly as quadratic expressions in the side vectors. The projection identities cause all mixed terms to cancel when the three corner areas are added, leaving exactly half the area of the original triangle. The geometric argument above is preferable because it keeps the role of the midpoint projections transparent and avoids substantial algebra.