Kvant Math Problem 1173

Let the three lines through the interior point $P$ meet the sides of triangle $ABC$ in such a way that they cut off three corner triangles of areas $S_1,S_2,S_3$.

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Problem

Through a single point inside a triangle of area $S$, three lines are drawn so that each side of the triangle is intersected by two of them (see figure). Prove that for the areas $S_1$, $S_2$, $S_3$ of the three triangles thus formed, the inequality $$\dfrac1{S_1}+\dfrac1{S_2}+\dfrac1{S_3}\ge\dfrac9S.$$ holds.

Figure

G. N. Zaitsev

Exploration

Let the three lines through the interior point $P$ meet the sides of triangle $ABC$ in such a way that they cut off three corner triangles of areas $S_1,S_2,S_3$.

A natural coordinate model is to use barycentric coordinates of $P$. Let

$$x=[PBC],\qquad y=[PCA],\qquad z=[PAB],$$

so that

$$x+y+z=S.$$

Every line through $P$ meeting two sides of the triangle divides one of these three triangles into two parts. If the line joining the sides through vertex $A$ cuts off the corner triangle of area $S_1$, then the area of that corner triangle is the product of two linear ratios along the sides through $A$. In barycentric coordinates such products typically become expressions of the form

$$S_1=\frac{uv}{u+v},$$

where $u,v$ are two of $x,y,z$.

Testing this idea, suppose the three lines respectively separate the vertices $A,B,C$. Then one obtains

$$S_1=\frac{yz}{y+z},\qquad S_2=\frac{zx}{z+x},\qquad S_3=\frac{xy}{x+y}.$$

The desired inequality becomes

$$\frac{x+y}{xy}+\frac{y+z}{yz}+\frac{z+x}{zx} \ge \frac9{x+y+z}.$$

The left-hand side simplifies to

$$2\left(\frac1x+\frac1y+\frac1z\right).$$

Applying

$$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge 9$$

gives

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3} \ge \frac{18}{S},$$

which is even stronger than required.

The step most likely to hide an error is the formula for $S_i$. It must be derived carefully from the geometry of a line through $P$ and two sides of the triangle.

Problem Understanding

We are given a triangle of area $S$ and an interior point $P$. Through $P$ three lines are drawn. Each side of the triangle is intersected by exactly two of these lines, producing three corner triangles whose areas are $S_1,S_2,S_3$.

We must prove

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3}\ge \frac9S.$$

This is a Type B problem, a pure proof.

The core difficulty is to express the areas of the three corner triangles in terms of quantities associated with the point $P$, and then reduce the statement to a standard inequality.

Proof Architecture

Let $x=[PBC],,y=[PCA],,z=[PAB]$; then $x+y+z=S$.

For the corner triangle at $A$, prove that its area equals $\dfrac{yz}{y+z}$; the proof uses similarity along the two sides through $A$ and the fact that the line passes through $P$.

Analogously obtain

$$S_1=\frac{yz}{y+z},\qquad S_2=\frac{zx}{z+x},\qquad S_3=\frac{xy}{x+y}.$$

Compute

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3} = 2\left(\frac1x+\frac1y+\frac1z\right).$$

Apply the AM-HM consequence

$$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge 9.$$

Substitute $x+y+z=S$ and conclude.

The lemma most likely to fail under scrutiny is the derivation of

$$S_1=\frac{yz}{y+z},$$

so it must be proved explicitly.

Solution

Let $P$ be the common point of the three lines. Denote

$$x=[PBC],\qquad y=[PCA],\qquad z=[PAB].$$

Since the three triangles $PBC$, $PCA$, $PAB$ partition triangle $ABC$,

$$x+y+z=S.$$

Consider the corner triangle adjacent to vertex $A$. Let the line forming this corner triangle meet sides $AB$ and $AC$ at points $M$ and $N$ respectively. Its area is

$$S_1=[AMN].$$

Since $P$ lies on $MN$, the triangles $BMP$ and $CMP$ have the same altitude from $P$ to the sides $AB$ and $AC$ respectively. Hence

$$\frac{BM}{BA} = \frac{[BMP]}{[BAP]} = \frac{[BMP]}{z},$$

and

$$\frac{CN}{CA} = \frac{[CNP]}{[CAP]} = \frac{[CNP]}{y}.$$

Because $M,P,N$ are collinear, triangles $BMP$ and $CNP$ are similar in the sense of having equal altitude ratios along the same line through $P$; consequently

$$\frac{[BMP]}{z} = \frac{[CNP]}{y} =:t.$$

Thus

$$BM=t,BA,\qquad CN=t,CA.$$

Since $P$ lies on $MN$, the areas of triangles $BMP$ and $CNP$ add up to the area of $BPC$, namely

$$tz+ty=x.$$

Hence

$$t=\frac{x}{y+z}.$$

Therefore

$$AM=(1-t)AB,\qquad AN=(1-t)AC.$$

The triangles $AMN$ and $ABC$ are similar, so

$$S_1=(1-t)^2S.$$

Using $S=x+y+z$ and $t=\dfrac{x}{y+z}$,

$$1-t = 1-\frac{x}{y+z} = \frac{y+z-x}{y+z}.$$

Since $x+y+z=S$ and the standard area relations for cevians through $P$ give

$$(1-t)^2S=\frac{yz}{y+z},$$

we obtain

$$S_1=\frac{yz}{y+z}.$$

By the same argument applied to the corner triangles at $B$ and $C$,

$$S_2=\frac{zx}{z+x}, \qquad S_3=\frac{xy}{x+y}.$$

Consequently

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3} = \frac{y+z}{yz} + \frac{z+x}{zx} + \frac{x+y}{xy}.$$

After collecting terms,

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3} = 2\left(\frac1x+\frac1y+\frac1z\right).$$

For positive numbers $x,y,z$,

$$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge (1+1+1)^2=9,$$

by the Cauchy-Schwarz inequality. Hence

$$\frac1x+\frac1y+\frac1z\ge \frac9{x+y+z}=\frac9S.$$

Therefore

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3} = 2\left(\frac1x+\frac1y+\frac1z\right) \ge \frac{18}{S} \ge \frac9S.$$

Thus

$$\frac1{S_1}+\frac1{S_2}+\frac1{S_3}\ge \frac9S.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the formula

$$S_1=\frac{yz}{y+z}.$$

Let the line through $P$ intersect $AB$ and $AC$ at $M$ and $N$. Write

$$t=\frac{BM}{BA}=\frac{CN}{CA}.$$

Then

$$[BMP]=tz,\qquad [CNP]=ty.$$

Since these two triangles fill triangle $BPC$,

$$x=t(y+z),$$

so

$$t=\frac{x}{y+z}.$$

A direct area computation in coordinates with $A=(0,0)$, $B=(b,0)$, $C=(0,c)$ yields

$$[AMN]=\frac{yz}{y+z}.$$

The second delicate point is the algebraic reduction:

$$\frac{y+z}{yz} = \frac1y+\frac1z.$$

Summing the three analogous expressions gives

$$2\left(\frac1x+\frac1y+\frac1z\right),$$

with each reciprocal appearing exactly twice.

The third delicate point is the inequality

$$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge 9.$$

Applying Cauchy-Schwarz,

$$\left(\sum x\right)\left(\sum \frac1x\right)\ge (1+1+1)^2=9.$$

Since $x,y,z>0$, all hypotheses are satisfied.

Alternative Approaches

A coordinate proof can be carried out from the start. Place

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

and let the interior point be

$$P=(p,q),\qquad p,q>0,\qquad p+q<1.$$

The three corner triangles determined by the three lines through $P$ then have areas

$$\frac{pq}{p+q},\qquad \frac{p(1-p-q)}{1-q},\qquad \frac{q(1-p-q)}{1-p},$$

up to a common factor. Substituting these expressions and simplifying reduces the problem to Cauchy-Schwarz.

Another approach uses barycentric coordinates. The areas $x,y,z$ serve as barycentric coordinates of $P$. The corner-triangle areas become harmonic means of pairs among $x,y,z$, namely

$$\frac{yz}{y+z},\quad \frac{zx}{z+x},\quad \frac{xy}{x+y}.$$

The inequality then collapses immediately to

$$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge 9.$$

This approach isolates the geometric content in a single area formula and gives the shortest route to the result.