Kvant Math Problem 1518
Let the common intersection point of the altitudes be $H$.
Verified: no
Verdicts: PASS + FAIL
Solve time: 23m19s
Source on kvant.digital
Problem
The altitudes of a tetrahedron intersect at a single point. Prove that this point, the foot of one of the altitudes, and the three points dividing the other altitudes in the ratio $2:1$, measured from the vertices, lie on the same sphere.
D. Teryoshin
Russian Mathematical Olympiad
Exploration
Let the common intersection point of the altitudes be $H$. Since the statement concerns a sphere through $H$, it is natural to place the origin of a vector coordinate system at $H$. Let the vertices of the tetrahedron be denoted by vectors
$$\vec A=a,\qquad \vec B=b,\qquad \vec C=c,\qquad \vec D=d.$$
Because the tetrahedron is orthocentric, the altitudes meet at $H$. For the altitude from $A$, the direction vector is $a$, and this altitude is perpendicular to the face $BCD$. Hence $a$ is orthogonal to every vector lying in the plane $BCD$, in particular
$$a\cdot(b-c)=0,\qquad a\cdot(c-d)=0.$$
Thus
$$a\cdot b=a\cdot c=a\cdot d.$$
Applying the same argument to the altitude from $B$ gives
$$b\cdot a=b\cdot c=b\cdot d.$$
Combining these equalities yields
$$a\cdot b=a\cdot c=a\cdot d=b\cdot c=b\cdot d=c\cdot d=:k.$$
The feet of the altitudes and the points dividing the altitudes can then be written explicitly. The problem reduces to constructing a sphere through the origin and four points lying on the rays $Ha$, $Hb$, $Hc$, and $Hd$.
Problem Understanding
Let $H_A$ be the foot of the altitude from $A$, and let $M_B$, $M_C$, $M_D$ divide the altitudes from $B$, $C$, $D$ in the ratio $2:1$ measured from the vertices. The goal is to prove that the points
$$H,\ H_A,\ M_B,\ M_C,\ M_D$$
lie on a single sphere. A convenient approach is to write an equation of a sphere passing through the origin. A sphere with center $m$ passing through the origin has the equation
$$|x|^2-2m\cdot x=0.$$
It suffices to construct a vector $m$ for which the four nontrivial points satisfy this equation.
Proof Architecture
Place the origin at $H$ and use the orthocentric tetrahedron identity
$$a\cdot b=a\cdot c=a\cdot d=b\cdot c=b\cdot d=c\cdot d=k.$$
Compute the vectors of $H_A$, $M_B$, $M_C$, and $M_D$. Establish the linear dependence among $a$, $b$, $c$, $d$ and derive a compatible system for the values $m\cdot a$, $m\cdot b$, $m\cdot c$, $m\cdot d$. After constructing $m$, verify directly that
$$|x|^2=2m\cdot x$$
for each of the four points. This will yield the sphere through all five points.
Solution
Place the origin at the orthocenter $H$ and write
$$a=\overrightarrow{HA},\qquad b=\overrightarrow{HB},\qquad c=\overrightarrow{HC},\qquad d=\overrightarrow{HD}.$$
As shown above,
$$a\cdot b=a\cdot c=a\cdot d=b\cdot c=b\cdot d=c\cdot d=k.$$
Let
$$\alpha=|a|^2,\qquad \beta=|b|^2,\qquad \gamma=|c|^2,\qquad \delta=|d|^2.$$
The plane $BCD$ consists of all points $x$ satisfying $a\cdot x=k$, because $b,c,d$ all satisfy this equation and the normal vector is $a$. Since $H_A$ lies on the line $\mathbb R a$, write $H_A=t a$. The condition $H_A\in BCD$ gives
$$a\cdot(ta)=t\alpha=k,$$
hence
$$H_A=\frac{k}{\alpha}a.$$
Similarly,
$$H_B=\frac{k}{\beta}b,\qquad H_C=\frac{k}{\gamma}c,\qquad H_D=\frac{k}{\delta}d.$$
Since $M_B$ divides $BH_B$ in the ratio $2:1$ measured from $B$,
$$M_B=\frac{B+2H_B}{3} =\frac{1}{3}\left(b+\frac{2k}{\beta}b\right) =\frac{\beta+2k}{3\beta},b.$$
Analogously,
$$M_C=\frac{\gamma+2k}{3\gamma},c, \qquad M_D=\frac{\delta+2k}{3\delta},d.$$
To justify the linear relation
$$\frac{a}{\alpha-k}+\frac{b}{\beta-k} +\frac{c}{\gamma-k}+\frac{d}{\delta-k}=0,$$
observe that $a,b,c,d\in\mathbb R^3$, so there is a nontrivial dependence
$$\lambda_A a+\lambda_B b+\lambda_C c+\lambda_D d=0.$$
Taking scalar products with $a,b,c,d$ gives
$$(\alpha-k)\lambda_A+k(\lambda_A+\lambda_B+\lambda_C+\lambda_D)=0,$$
$$(\beta-k)\lambda_B+k(\lambda_A+\lambda_B+\lambda_C+\lambda_D)=0,$$
$$(\gamma-k)\lambda_C+k(\lambda_A+\lambda_B+\lambda_C+\lambda_D)=0,$$
$$(\delta-k)\lambda_D+k(\lambda_A+\lambda_B+\lambda_C+\lambda_D)=0.$$
Subtracting the first equation from the other three yields
$$(\alpha-k)\lambda_A=(\beta-k)\lambda_B =(\gamma-k)\lambda_C =(\delta-k)\lambda_D.$$
Let the common value be $t$. Then
$$\lambda_A=\frac{t}{\alpha-k},\qquad \lambda_B=\frac{t}{\beta-k},\qquad \lambda_C=\frac{t}{\gamma-k},\qquad \lambda_D=\frac{t}{\delta-k}.$$
If $t=0$, all $\lambda$'s would vanish, contradicting nontriviality. Hence $t\neq0$, and dividing the dependence relation by $t$ gives
$$\frac{a}{\alpha-k}+\frac{b}{\beta-k} +\frac{c}{\gamma-k}+\frac{d}{\delta-k}=0.$$
Define
$$r_A=\frac{k}{2},\qquad r_B=\frac{\beta+2k}{6},\qquad r_C=\frac{\gamma+2k}{6},\qquad r_D=\frac{\delta+2k}{6}.$$
We compute
$$\frac{r_A}{\alpha-k} +\frac{r_B}{\beta-k} +\frac{r_C}{\gamma-k} +\frac{r_D}{\delta-k}.$$
Using
$$\frac{\beta+2k}{\beta-k} =1+\frac{3k}{\beta-k},$$
and the analogous formulas for $\gamma,\delta$, this sum equals
$$\frac16\left( 3+3k\left( \frac1{\alpha-k} +\frac1{\beta-k} +\frac1{\gamma-k} +\frac1{\delta-k} \right) \right).$$
Taking the scalar product of
$$\frac{a}{\alpha-k}+\frac{b}{\beta-k} +\frac{c}{\gamma-k}+\frac{d}{\delta-k}=0$$
with $a$ gives
$$\frac{\alpha}{\alpha-k} +\frac{k}{\beta-k} +\frac{k}{\gamma-k} +\frac{k}{\delta-k}=0.$$
Since
$$\frac{\alpha}{\alpha-k} =1+\frac{k}{\alpha-k},$$
we obtain
$$1+k\left( \frac1{\alpha-k} +\frac1{\beta-k} +\frac1{\gamma-k} +\frac1{\delta-k} \right)=0.$$
Substituting this into the previous expression shows that
$$\frac{r_A}{\alpha-k} +\frac{r_B}{\beta-k} +\frac{r_C}{\gamma-k} +\frac{r_D}{\delta-k}=0.$$
Thus the numbers $r_A,r_B,r_C,r_D$ satisfy the same linear relation as the coefficients of $a,b,c,d$.
Consider the linear map
$$T:\mathbb R^3\to\mathbb R^4,\qquad T(m)=(m\cdot a,m\cdot b,m\cdot c,m\cdot d).$$
The image of $T$ consists precisely of those quadruples $(s_A,s_B,s_C,s_D)$ satisfying every linear relation among $a,b,c,d$. We have proved that $(r_A,r_B,r_C,r_D)$ satisfies the relation
$$\frac{s_A}{\alpha-k} +\frac{s_B}{\beta-k} +\frac{s_C}{\gamma-k} +\frac{s_D}{\delta-k}=0,$$
which is the unique relation among $a,b,c,d$ up to a nonzero factor. Hence $(r_A,r_B,r_C,r_D)$ belongs to the image of $T$, so there exists a vector $m$ such that
$$m\cdot a=r_A,\qquad m\cdot b=r_B,\qquad m\cdot c=r_C,\qquad m\cdot d=r_D.$$
We now verify the sphere equation.
For $H_A=\dfrac{k}{\alpha}a$,
$$|H_A|^2 =\left(\frac{k}{\alpha}\right)^2|a|^2 =\frac{k^2}{\alpha},$$
while
$$2m\cdot H_A =2\frac{k}{\alpha}(m\cdot a) =2\frac{k}{\alpha}\cdot\frac{k}{2} =\frac{k^2}{\alpha}.$$
Hence
$$|H_A|^2=2m\cdot H_A.$$
For
$$M_B=t_B b,\qquad t_B=\frac{\beta+2k}{3\beta},$$
we have
$$|M_B|^2=t_B^2\beta =\frac{(\beta+2k)^2}{9\beta},$$
and
$$2m\cdot M_B =2t_B(m\cdot b) =2\frac{\beta+2k}{3\beta}\cdot\frac{\beta+2k}{6} =\frac{(\beta+2k)^2}{9\beta}.$$
Thus
$$|M_B|^2=2m\cdot M_B.$$
The same calculation gives
$$|M_C|^2=2m\cdot M_C, \qquad |M_D|^2=2m\cdot M_D.$$
Therefore all four points $H_A,M_B,M_C,M_D$ satisfy
$$|x|^2-2m\cdot x=0.$$
The origin $H$ also satisfies this equation. Hence $H,H_A,M_B,M_C,M_D$ all lie on the sphere with center $m$ and equation
$$|x|^2-2m\cdot x=0.$$
This proves that the five points are concyclic on a common sphere.