Kvant Math Problem 843
Let the plane of the triangle be $z=0$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 10m37s
Source on kvant.digital
Problem
At the vertices of triangle $ABC$, perpendiculars $AA_1$, $BB_1$, and $CC_1$ to its plane are erected on the same side of the plane, each having length equal to the corresponding altitude of the triangle. Prove that the perpendicular dropped from the point of intersection of the planes $ABC_1$, $BCA_1$, and $CAB_1$ onto the plane $ABC$ lands at the center of the circle inscribed in triangle $ABC$ and has length equal to its radius.
A. A. Yagubyants
Exploration
Let the plane of the triangle be $z=0$. Denote by $a,b,c$ the side lengths opposite $A,B,C$, and by $h_a,h_b,h_c$ the corresponding altitudes. Since
$$h_a=\frac{2\Delta}{a},\qquad h_b=\frac{2\Delta}{b},\qquad h_c=\frac{2\Delta}{c},$$
where $\Delta$ is the area of $ABC$, the lifted points are
$$A_1=(A,h_a),\qquad B_1=(B,h_b),\qquad C_1=(C,h_c).$$
The three planes are $ABC_1$, $BCA_1$, $CAB_1$. Their common point $P$ should somehow encode the three altitudes. The statement suggests that the projection of $P$ onto $ABC$ is the incenter $I$, and that the height of $P$ above the plane equals the inradius
$$r=\frac{\Delta}{s},$$
where $s=\frac{a+b+c}{2}$.
A natural coordinate description of a point above the plane is obtained through barycentric coordinates. If $Q$ projects to a point with barycentric coordinates $(\alpha,\beta,\gamma)$ relative to $ABC$, then a point of the plane $ABC_1$ has height equal to $\gamma h_c$, because on that plane the height varies linearly, being $0$ on $AB$ and $h_c$ at $C$. Similarly, on $BCA_1$ the height is $\alpha h_a$, and on $CAB_1$ it is $\beta h_b$.
Hence a common point of the three planes must satisfy
$$\alpha h_a=\beta h_b=\gamma h_c.$$
Substituting $h_a:h_b:h_c=1/a:1/b:1/c$ gives
$$\frac{\alpha}{a}=\frac{\beta}{b}=\frac{\gamma}{c}.$$
Therefore
$$(\alpha,\beta,\gamma)\propto(a,b,c),$$
which are exactly the barycentric coordinates of the incenter. This identifies the projection.
The remaining task is to compute the common height. If $(\alpha,\beta,\gamma)=\left(\frac a{2s},\frac b{2s},\frac c{2s}\right)$, then
$$z=\alpha h_a =\frac a{2s}\cdot\frac{2\Delta}{a} =\frac{\Delta}{s} =r.$$
The potentially dangerous step is the assertion that the equation of the plane $ABC_1$ is $z=\gamma h_c$ in barycentric coordinates. That must be proved carefully.
Problem Understanding
We are given a triangle $ABC$. At each vertex a segment perpendicular to the plane of the triangle is erected on the same side, with lengths equal to the corresponding altitudes of the triangle. Their endpoints are $A_1,B_1,C_1$.
The planes $ABC_1$, $BCA_1$, and $CAB_1$ intersect at a point $P$. We must prove that the orthogonal projection of $P$ onto the plane of $ABC$ is the incenter of the triangle, and that the distance from $P$ to the plane equals the inradius.
This is a Type B problem. The core difficulty is to translate membership in the three planes into a condition on the barycentric coordinates of the projection of $P$ onto the triangle's plane.
Proof Architecture
First, introduce coordinates with the plane $ABC$ given by $z=0$, and represent every point by its projection onto $ABC$ together with its height.
Second, prove that if a point projects to barycentric coordinates $(\alpha,\beta,\gamma)$, then its height on the plane $ABC_1$ equals $\gamma h_c$; this follows because height is an affine function on the plane and takes values $0,0,h_c$ at $A,B,C$.
Third, derive analogous formulas for the planes $BCA_1$ and $CAB_1$, obtaining heights $\alpha h_a$ and $\beta h_b$.
Fourth, use the existence of the common point $P$ to obtain
$$\alpha h_a=\beta h_b=\gamma h_c.$$
Substituting $h_a=2\Delta/a$, $h_b=2\Delta/b$, $h_c=2\Delta/c$ yields
$$\alpha:a=\beta:b=\gamma:c.$$
Hence the projection of $P$ has barycentric coordinates proportional to $(a,b,c)$, so it is the incenter.
Finally, compute the common height and show that it equals
$$\frac{\Delta}{s}=r.$$
The lemma most likely to fail under scrutiny is the affine-height description of the plane $ABC_1$.
Solution
Let the plane of triangle $ABC$ be the plane $z=0$. Denote by
$$h_a,\quad h_b,\quad h_c$$
the altitudes of triangle $ABC$ drawn from $A,B,C$ respectively. Then
$$A_1=(A,h_a),\qquad B_1=(B,h_b),\qquad C_1=(C,h_c).$$
Let $P$ be the common point of the planes $ABC_1$, $BCA_1$, and $CAB_1$. Denote by $X$ the orthogonal projection of $P$ onto the plane $ABC$.
Write the barycentric coordinates of $X$ with respect to triangle $ABC$ as
$$(\alpha,\beta,\gamma), \qquad \alpha+\beta+\gamma=1.$$
Let $z(P)$ denote the height of $P$ above the plane $ABC$.
Consider first the plane $ABC_1$. On this plane the height above $ABC$ is an affine function of position. At the vertices $A,B,C$ its values are
$$0,\quad 0,\quad h_c.$$
Hence, for a point whose projection onto $ABC$ has barycentric coordinates $(\alpha,\beta,\gamma)$, the height on the plane $ABC_1$ equals
$$\alpha\cdot0+\beta\cdot0+\gamma h_c = \gamma h_c.$$
Since $P\in ABC_1$,
$$z(P)=\gamma h_c.$$
Applying the same argument to the plane $BCA_1$, whose heights at $A,B,C$ are $h_a,0,0$, gives
$$z(P)=\alpha h_a.$$
For the plane $CAB_1$, whose heights at $A,B,C$ are $0,h_b,0$, we obtain
$$z(P)=\beta h_b.$$
Therefore
$$\alpha h_a=\beta h_b=\gamma h_c.$$
Let this common value be $t$. Since
$$h_a=\frac{2\Delta}{a},\qquad h_b=\frac{2\Delta}{b},\qquad h_c=\frac{2\Delta}{c},$$
where $\Delta=[ABC]$, we get
$$\frac{2\Delta}{a}\alpha = \frac{2\Delta}{b}\beta = \frac{2\Delta}{c}\gamma.$$
After cancelling $2\Delta$,
$$\frac{\alpha}{a} = \frac{\beta}{b} = \frac{\gamma}{c}.$$
Thus
$$(\alpha,\beta,\gamma)\propto(a,b,c).$$
The barycentric coordinates of the incenter of a triangle are proportional to the side lengths $(a,b,c)$. Hence $X$ is the incenter of triangle $ABC$.
Let
$$s=\frac{a+b+c}{2}.$$
Since $\alpha+\beta+\gamma=1$ and $(\alpha,\beta,\gamma)\propto(a,b,c)$,
$$\alpha=\frac{a}{2s},\qquad \beta=\frac{b}{2s},\qquad \gamma=\frac{c}{2s}.$$
Using $z(P)=\alpha h_a$,
$$z(P) = \frac{a}{2s}\cdot\frac{2\Delta}{a} = \frac{\Delta}{s}.$$
But
$$r=\frac{\Delta}{s}$$
is the inradius of triangle $ABC$. Therefore
$$z(P)=r.$$
Thus the perpendicular from $P$ to the plane $ABC$ meets that plane at the incenter of triangle $ABC$, and its length equals the inradius.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the formula $z=\gamma h_c$ on the plane $ABC_1$. Let a point of that plane be written as
$$Y=\alpha A+\beta B+\gamma C_1, \qquad \alpha+\beta+\gamma=1.$$
Its projection onto $ABC$ is
$$\alpha A+\beta B+\gamma C,$$
which has barycentric coordinates $(\alpha,\beta,\gamma)$. Its height equals the coefficient of the vertical component of $C_1$, namely $\gamma h_c$. This independently confirms the affine-height argument.
The second delicate step is the identification of the incenter. From
$$\frac{\alpha}{a} = \frac{\beta}{b} = \frac{\gamma}{c}$$
there exists $\lambda$ such that
$$\alpha=\lambda a,\qquad \beta=\lambda b,\qquad \gamma=\lambda c.$$
Since $\alpha+\beta+\gamma=1$,
$$\lambda=\frac1{a+b+c}.$$
Hence the barycentric coordinates are exactly proportional to $(a,b,c)$, which is the standard characterization of the incenter.
The third delicate step is the computation of the height. Substituting
$$\alpha=\frac{a}{a+b+c}$$
and
$$h_a=\frac{2\Delta}{a}$$
gives
$$z(P)=\frac{2\Delta}{a+b+c}.$$
Since
$$s=\frac{a+b+c}{2},$$
this becomes
$$z(P)=\frac{\Delta}{s}=r.$$
No hidden normalization factor is lost.
Alternative Approaches
A different proof can be written in trilinear coordinates. The planes $ABC_1$, $BCA_1$, and $CAB_1$ determine a point whose projection $X$ satisfies
$$\frac{d(X,BC)}{h_a} = \frac{d(X,CA)}{h_b} = \frac{d(X,AB)}{h_c}.$$
Since $h_a:h_b:h_c=1/a:1/b:1/c$, this becomes
$$d(X,BC)=d(X,CA)=d(X,AB),$$
which characterizes the incenter directly. The common height is then found from any one of the plane equations.
The barycentric approach is preferable because it converts each plane into a simple linear height function. The three plane conditions immediately yield the proportionality $(\alpha,\beta,\gamma)\propto(a,b,c)$ and the height computation follows in one line.