IMO 1981 Problem 1
Let $a=BC$, $b=CA$, $c=AB$.
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 13m46s
Problem
$\displaystyle P$ is a point inside a given triangle $\displaystyle ABC$. $\displaystyle D, E, F$ are the feet of the perpendiculars from $\displaystyle P$ to the lines $\displaystyle BC, CA, AB$, respectively. Find all $\displaystyle P$ for which
$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$
is least.
Exploration
Let $a=BC$, $b=CA$, $c=AB$. The expression becomes
$$S(P)=\frac{a}{PD}+\frac{b}{PE}+\frac{c}{PF}.$$
Using area decompositions, each perpendicular distance can be rewritten through triangle areas:
$$[ PBC ]=\frac{1}{2}a\cdot PD,\quad [ PCA ]=\frac{1}{2}b\cdot PE,\quad [ PAB ]=\frac{1}{2}c\cdot PF.$$
Hence each term has the form $a/PD = a^2/(2[ PBC ])$, and similarly for the others. The constraint $[PBC]+[PCA]+[PAB]=[ABC]$ is fixed, so the problem becomes a constrained minimization with reciprocal area terms.
A natural candidate for extremality is when the three subareas are proportional to $a,b,c$, since this aligns the symmetry of numerators and denominators in a Cauchy-type inequality. This configuration corresponds to the incenter of the triangle.
The key risk is misapplying Cauchy-Schwarz without tracking equality conditions and ensuring geometric realizability of the resulting area ratios.
Problem Understanding
This is a Type C problem, requiring the minimum value of a geometric expression depending on a point $P$ inside triangle $ABC$.
The expression sums side lengths divided by distances from $P$ to the corresponding sides. The goal is to determine the position of $P$ that minimizes this sum and compute the minimum value.
The main difficulty lies in converting geometric distances into a global constraint that allows a clean inequality, and then identifying when equality is geometrically meaningful.
The expected extremum occurs when $P$ is the incenter of $ABC$, since it equalizes distances to the sides and matches the symmetry of the triangle’s side lengths in a variational balance.
Proof Architecture
Let $a=BC$, $b=CA$, $c=AB$.
First, a lemma expresses each term $a/PD$, $b/PE$, $c/PF$ in terms of areas $[PBC]$, $[PCA]$, $[PAB]$.
Second, a lemma applies the Engel form of Cauchy-Schwarz to the resulting expression to produce a lower bound depending only on $a,b,c$ and $[ABC]$.
Third, a lemma characterizes the equality condition as $[PBC]:[PCA]:[PAB]=a:b:c$.
Fourth, a final lemma identifies this barycentric condition with $P$ being the incenter.
The most delicate step is ensuring that the equality condition corresponds to a point inside the triangle and not merely a formal ratio condition.
Solution
Let $a=BC$, $b=CA$, $c=AB$, and let $[XYZ]$ denote the area of triangle $XYZ$. Let $K=[ABC]$.
Lemma 1
For any interior point $P$ with feet $D,E,F$ on $BC,CA,AB$ respectively,
$$\frac{a}{PD}=\frac{a^2}{2[ PBC ]},\quad \frac{b}{PE}=\frac{b^2}{2[ PCA ]},\quad \frac{c}{PF}=\frac{c^2}{2[ PAB ]}.$$
From $[PBC]=\frac{1}{2}a\cdot PD$, rearrangement yields $PD=\frac{2[ PBC ]}{a}$ and substitution gives the identity, with analogous computations for the other sides. This establishes a direct translation of the geometric expression into area variables, preventing dependence on local perpendicular distances.
Lemma 2
For positive $x_1,x_2,x_3$ with $x_1+x_2+x_3=K$,
$$\frac{a^2}{2x_1}+\frac{b^2}{2x_2}+\frac{c^2}{2x_3}\ge \frac{(a+b+c)^2}{2K}.$$
Applying Cauchy-Schwarz in Engel form,
$$\frac{a^2}{x_1}+\frac{b^2}{x_2}+\frac{c^2}{x_3} \ge \frac{(a+b+c)^2}{x_1+x_2+x_3} =\frac{(a+b+c)^2}{K},$$
and division by $2$ gives the claim. This step isolates the dependence on $P$ entirely through a fixed sum constraint, eliminating geometric variability.
Lemma 3
Equality holds in Lemma 2 if and only if
$$[ PBC ]:[ PCA ]:[ PAB ]=a:b:c.$$
Equality in Cauchy-Schwarz requires proportionality of sequences, hence $\frac{a}{[PBC]}=\frac{b}{[PCA]}=\frac{c}{[PAB]}$, which is equivalent to the stated ratio condition. This identifies the unique structural configuration achieving extremality.
Lemma 4
A point $P$ inside triangle $ABC$ satisfies
$$[ PBC ]:[ PCA ]:[ PAB ]=a:b:c$$
if and only if $P$ is the incenter of $ABC$.
This follows because the incenter is characterized by equal distances to the sides, say $r$, yielding
$$[ PBC ]=\frac{1}{2}ar,\quad [ PCA ]=\frac{1}{2}br,\quad [ PAB ]=\frac{1}{2}cr,$$
which produces the required proportionality. Conversely, the ratio condition determines equal angles at $P$ with the sides, forcing $P$ to be the intersection of angle bisectors, hence the incenter.
Completion of the argument
Combining Lemma 1 and Lemma 2 yields
$$S(P)=\frac{a}{PD}+\frac{b}{PE}+\frac{c}{PF} =\frac{a^2}{2[ PBC ]}+\frac{b^2}{2[ PCA ]}+\frac{c^2}{2[ PAB ]} \ge \frac{(a+b+c)^2}{2K}.$$
Equality holds precisely at the incenter $I$. For the incenter, $PD=PE=PF=r$ and $K=rs$ where $s=\frac{a+b+c}{2}$. Therefore,
$$S(I)=\frac{a+b+c}{r}=\frac{(a+b+c)^2}{2K}.$$
The minimum value is achieved at the incenter.
$$\boxed{\frac{a+b+c}{r}}$$
Verification of Key Steps
The area substitution relies only on the standard formula for triangle area with a chosen base and corresponding height, ensuring no hidden geometric assumptions.
The inequality step uses Cauchy-Schwarz in Engel form with strict positivity of areas, and equality conditions require proportionality of numerators and denominators, which has been explicitly checked.
The identification of the incenter follows from the characterization of points equidistant in weighted area ratios, and no alternative interior point satisfies the same proportionality without coinciding angle bisectors.
Alternative Approaches
A synthetic approach replaces area inequalities with barycentric coordinates, writing the expression directly in terms of coordinates proportional to subtriangle areas and minimizing a convex function on the simplex. Another approach uses Lagrange multipliers on the constraint $[PBC]+[PCA]+[PAB]=K$, leading to the same proportionality condition. The area method is preferable because it isolates the geometry into a single fixed constraint and reduces the problem to a direct application of a sharp inequality with transparent equality conditions.