Kvant Math Problem 716
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Problem
From a point $P$ inside a given triangle $ABC$, perpendiculars $PA_1$, $PB_1$, $PC_1$ are dropped to the lines $BC$, $CA$, and $AB$. For which points $P$ inside $\triangle ABC$ does the quantity $$\dfrac{|BC|}{|PA_1|}+\dfrac{|CA|}{|PB_1|}+\dfrac{|AB|}{|PC_1|}$$ attain its minimum value?
International Mathematical Olympiad for School Students (XXII, 1981)
Exploration
Let
$$x=|PA_1|,\qquad y=|PB_1|,\qquad z=|PC_1|.$$
These are the distances from $P$ to the three sides of the triangle. The quantity to minimize is
$$S=\frac{a}{x}+\frac{b}{y}+\frac{c}{z},$$
where
$$a=|BC|,\qquad b=|CA|,\qquad c=|AB|.$$
The distances are not independent. Since $P$ lies inside the triangle, the areas of triangles $PBC$, $PCA$, $PAB$ add up to the area $\Delta$ of $ABC$:
$$\Delta=\frac12(ax+by+cz).$$
Hence
$$ax+by+cz=2\Delta,$$
a fixed constant.
The problem has become: minimize
$$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$$
subject to
$$ax+by+cz=2\Delta,\qquad x,y,z>0.$$
A natural guess is that equality in some convexity inequality will force $x=y=z$. If that happens, then $P$ is equidistant from all three sides, hence $P$ is the incenter.
To test the plausibility, consider the equilateral case. Then the incenter has $x=y=z=h/3$, and symmetry strongly suggests the minimum occurs there. If one distance is made smaller while keeping $ax+by+cz$ fixed, the reciprocal term grows rapidly. This supports the conjecture.
The crucial point is proving the lower bound rigorously from the linear constraint. The most direct route is Cauchy-Schwarz:
$$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)(ax+by+cz) \ge (a+b+c)^2.$$
Since the second factor is fixed, this immediately yields a lower bound, and equality requires
$$x=y=z.$$
That identifies the minimizing point as the incenter.
Problem Understanding
We are given a triangle $ABC$ and an interior point $P$. Let $x,y,z$ denote the perpendicular distances from $P$ to the sides $BC,CA,AB$. We must determine for which interior points $P$ the quantity
$$\frac{|BC|}{x}+\frac{|CA|}{y}+\frac{|AB|}{z}$$
is minimal.
This is a Type C problem. We must determine the minimum and characterize all points where it is attained.
The core difficulty is converting the geometric condition into an algebraic constraint on the distances $x,y,z$ and then proving a sharp lower bound for the reciprocal expression.
The answer should be the incenter of the triangle. The reason is that the distances from the incenter to the three sides are equal, and the area identity imposes a fixed weighted sum $ax+by+cz$, making equality in a suitable inequality possible only when $x=y=z$.
Proof Architecture
Let $a=|BC|$, $b=|CA|$, $c=|AB|$, and let $\Delta$ be the area of triangle $ABC$.
First claim: the distances $x,y,z$ satisfy
$$ax+by+cz=2\Delta.$$
This follows by decomposing the area of $ABC$ into the areas of triangles $PBC$, $PCA$, and $PAB$.
Second claim: for positive $x,y,z$,
$$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)(ax+by+cz)\ge(a+b+c)^2.$$
This is an application of the Cauchy-Schwarz inequality.
Third claim: equality holds if and only if
$$x=y=z.$$
This is the equality condition in Cauchy-Schwarz.
Fourth claim: an interior point has equal distances to the three sides if and only if it is the incenter.
This is the standard characterization of the intersection of the three internal angle bisectors.
The hardest step is establishing the sharp lower bound together with the exact equality condition.
Solution
Denote
$$a=|BC|,\qquad b=|CA|,\qquad c=|AB|,$$
and let
$$x=|PA_1|,\qquad y=|PB_1|,\qquad z=|PC_1|.$$
Let $\Delta$ be the area of triangle $ABC$.
Since $P$ lies inside the triangle, the triangles $PBC$, $PCA$, and $PAB$ form a partition of $ABC$. Hence
$$\Delta=[PBC]+[PCA]+[PAB].$$
Using the usual area formula,
$$[PBC]=\frac12 ax,\qquad [PCA]=\frac12 by,\qquad [PAB]=\frac12 cz.$$
Therefore
$$\Delta=\frac12(ax+by+cz),$$
or
$$ax+by+cz=2\Delta.$$
The quantity to be minimized is
$$S=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}.$$
Apply the Cauchy-Schwarz inequality in the form
$$\left(\sum \frac{a}{x}\right)\left(\sum ax\right) \ge (a+b+c)^2.$$
Thus
$$S(ax+by+cz)\ge(a+b+c)^2.$$
Using $ax+by+cz=2\Delta$,
$$S\ge\frac{(a+b+c)^2}{2\Delta}.$$
Hence
$$S \ge \frac{(a+b+c)^2}{2\Delta}.$$
This lower bound is independent of $P$.
It remains to determine when equality occurs. Equality in Cauchy-Schwarz holds precisely when
$$\frac1x=\frac1y=\frac1z,$$
equivalently,
$$x=y=z.$$
Therefore equality holds exactly when the distances from $P$ to the three sides of the triangle are equal.
A point inside a triangle is equidistant from the sides $AB$ and $AC$ if and only if it lies on the internal bisector of angle $A$. Similarly, equality of the distances to the sides adjacent to vertices $B$ and $C$ places the point on the corresponding internal bisectors. Hence a point equidistant from all three sides is the common point of the internal angle bisectors, namely the incenter.
For the incenter $I$, the common distance to the sides equals the inradius $r$. Then
$$S=\frac{a+b+c}{r}.$$
Since
$$\Delta=rs, \qquad s=\frac{a+b+c}{2},$$
we obtain
$$S=\frac{a+b+c}{r} =\frac{(a+b+c)^2}{2\Delta},$$
which is exactly the lower bound.
Therefore the minimum value is
$$\boxed{\frac{(|AB|+|BC|+|CA|)^2}{2[ABC]}}$$
and equality holds only at the incenter of the triangle.
Verification of Key Steps
The first delicate step is the relation
$$ax+by+cz=2\Delta.$$
The argument uses only area decomposition. Since $P$ is interior, the three triangles $PBC$, $PCA$, and $PAB$ are disjoint and cover $ABC$. Writing each area as one half times a side length times the corresponding perpendicular distance gives the identity exactly. If $P$ were outside the triangle, signed areas would be required and the formula would need modification.
The second delicate step is the Cauchy-Schwarz application. Taking
$$u_i=\sqrt{\frac{a_i}{x_i}}, \qquad v_i=\sqrt{a_i x_i},$$
with $(a_1,a_2,a_3)=(a,b,c)$ and $(x_1,x_2,x_3)=(x,y,z)$, Cauchy-Schwarz gives
$$\left(\sum \frac{a_i}{x_i}\right) \left(\sum a_i x_i\right) \ge \left(\sum a_i\right)^2.$$
The equality condition is
$$u_i=\lambda v_i \quad\text{for all }i,$$
which becomes
$$\frac1{x_i}=\lambda x_i.$$
Since all $x_i$ are positive, this implies
$$x_1=x_2=x_3.$$
The third delicate step is the characterization of the incenter. The locus of points equidistant from the sides $AB$ and $AC$ is the internal bisector of angle $A$. Thus a point satisfying $x=y=z$ lies simultaneously on the bisectors of angles $A$ and $B$, hence is their intersection, the incenter. No other interior point has this property.
Alternative Approaches
A different solution uses the arithmetic mean-harmonic mean inequality. Set
$$u=ax,\qquad v=by,\qquad w=cz.$$
Then
$$u+v+w=2\Delta$$
and
$$S=\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}.$$
Applying Engel's form of Cauchy,
$$\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w} \ge \frac{(a+b+c)^2}{u+v+w} = \frac{(a+b+c)^2}{2\Delta}.$$
Equality requires
$$\frac{u}{a}=\frac{v}{b}=\frac{w}{c},$$
which again yields $x=y=z$ and hence the incenter.
The main approach is preferable because it works directly with the original expression and the area identity, producing the minimum and its equality condition in a single application of Cauchy-Schwarz.