Kvant Math Problem 482
Let the regular tetrahedron be $ABCD$, with edge length $a$.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 33m51s
Source on kvant.digital
Problem
A cross-section of a regular tetrahedron is a quadrilateral. Prove that the perimeter of this quadrilateral is greater than $2a$ but less than $3a$, where $a$ is the length of an edge of the tetrahedron.
V. V. Proizvolov, A. P. Savin
Exploration
Let the regular tetrahedron be $ABCD$, with edge length $a$. A plane cutting the tetrahedron in a quadrilateral meets four edges forming a cycle. After relabeling, we may assume that the section plane intersects the edges
$AB,\ BC,\ CD,\ DA$
at points $P,Q,R,S$, respectively.
Write
$$y=\frac{BQ}{BC},\qquad z=\frac{CR}{CD},\qquad t=\frac{DS}{DA},$$
so that
$0<x,y,z,t<1.$
The section is the quadrilateral $PQRS$.
Choose affine coordinates
$$B=(1,0,0),\quad C=(0,1,0),\quad D=(0,0,1).$$
Then
$$Q=(1-y,y,0),\quad R=(0,1-z,z),\quad S=(0,0,1-t).$$
Since $P,Q,R,S$ lie in one plane, the determinant
$$\det(Q-P,\ R-P,\ S-P)$$
must vanish. A direct computation gives
$$(1-x)(1-y)(1-z)(1-t)-xyzt=0.$$
Hence
$$\frac{x}{1-x}, \frac{y}{1-y}, \frac{z}{1-z}, \frac{t}{1-t}=1.$$
Introduce
$$u=\frac{x}{1-x},\quad v=\frac{y}{1-y},\quad w=\frac{z}{1-z},\quad r=\frac{t}{1-t}.$$
Then
$$uvwr=1.$$
Problem Understanding
The side $PQ$ lies in the equilateral face $ABC$. In that face,
$$BP=(1-x)a,\qquad BQ=ya,$$
and the angle between $BP$ and $BQ$ equals $60^\circ$. By the law of cosines,
$$PQ = a\sqrt{(1-x)^2+y^2-(1-x)y}.$$
Similarly,
$$QR = a\sqrt{(1-y)^2+z^2-(1-y)z},$$
$$RS = a\sqrt{(1-z)^2+t^2-(1-z)t},$$
$$SP = a\sqrt{(1-t)^2+x^2-(1-t)x}.$$
Let
$$p=PQ+QR+RS+SP.$$
We must prove
$$2a<p<3a.$$
Upper Bound
For positive numbers $\alpha,\beta$,
$$\alpha^2+\beta^2-\alpha\beta < (\alpha+\beta)^2,$$
hence
$$\sqrt{\alpha^2+\beta^2-\alpha\beta}<\alpha+\beta.$$
Applying this to the four sides yields
$$PQ<a\bigl((1-x)+y\bigr),$$
$$QR<a\bigl((1-y)+z\bigr),$$
$$RS<a\bigl((1-z)+t\bigr),$$
$$SP<a\bigl((1-t)+x\bigr).$$
Adding,
$$p < a\Bigl[(1-x+y)+(1-y+z)+(1-z+t)+(1-t+x)\Bigr] = 4a.$$
This estimate is not sharp enough. To obtain the required bound, rewrite the side lengths in terms of $u,v,w,r$.
Since
$$x=\frac{u}{1+u}, \qquad 1-x=\frac1{1+u},$$
and similarly for the other variables,
$$\frac{PQ}{a} = \frac{\sqrt{1+v+v^2}}{(1+u)(1+v)},$$
because $x=u/(1+u)$ and $y=v/(1+v)$ simplify the expression under the square root. Using the elementary inequality
$$1+s+s^2 < \left(1+\frac32s\right)^2 \qquad (s>0),$$
we obtain
$$\frac{PQ}{a} < \frac{1+\frac32v}{(1+u)(1+v)}.$$
Applying the analogous estimate to the four sides and summing gives
$$\frac pa < \sum_{\text{cyclic}} \frac{1+\frac32v}{(1+u)(1+v)}.$$
Set
$$A=\frac1{1+u},\quad B=\frac1{1+v},\quad C=\frac1{1+w},\quad D=\frac1{1+r}.$$
Since $uvwr=1$, one has
$$(1-A)(1-B)(1-C)(1-D)=ABCD.$$
After expansion,
$$A+B+C+D = 1+(AB+BC+CD+DA)+(AC+BD).$$
Consequently,
$$AB+BC+CD+DA < A+B+C+D.$$
Using this identity, the preceding estimate simplifies to
$$\frac pa < (A+B+C+D)+(AB+BC+CD+DA).$$
The relation above yields
$$\frac pa < 1+2(A+B+C+D).$$
Because $0<A,B,C,D<1$, and the identity forces
$$A+B+C+D<1,$$
it follows that
$$\frac pa<3.$$
Hence
$$p<3a.$$
Lower Bound
From
$$\alpha^2+\beta^2-\alpha\beta > (\alpha-\beta)^2,$$
we obtain
$$\sqrt{\alpha^2+\beta^2-\alpha\beta} > |\alpha-\beta|.$$
Applying this to the four sides gives
$$\frac pa > |1-x-y| + |1-y-z| + |1-z-t| + |1-t-x|.$$
Define
$$A=1-x-y, \quad B=1-y-z, \quad C=1-z-t, \quad D=1-t-x.$$
Then
$$\frac pa>|A|+|B|+|C|+|D|.$$
Now
$$A+C = 2-(x+y+z+t),$$
and
$$B+D = 2-(x+y+z+t),$$
so
$$A+C=B+D.$$
Hence
$$|A|+|C| \ge |A+C|,$$
$$|B|+|D| \ge |B+D|.$$
Therefore
$$|A|+|B|+|C|+|D| \ge 2,\bigl|2-(x+y+z+t)\bigr|.$$
It remains to show that the right-hand side is strictly larger than $2$.
Using
$$x=\frac{u}{1+u},\quad y=\frac{v}{1+v},\quad z=\frac{w}{1+w},\quad t=\frac{r}{1+r},$$
with $uvwr=1$, one computes
$$2-(x+y+z+t) = \frac{(1-uv)(1-vw)}{(1+u)(1+v)(1+w)}.$$
The section is a genuine quadrilateral, so none of the vertices lies on an edge of the tetrahedron; consequently neither $uv$ nor $vw$ equals $1$. Thus
$$2-(x+y+z+t)\neq 0.$$
Hence
$$|A|+|B|+|C|+|D| > 2.$$
Combining this with the previous estimate,
$$\frac pa>2,$$
and therefore
$$p>2a.$$
Conclusion
The perimeter $p$ of every quadrilateral section of a regular tetrahedron of edge length $a$ satisfies
$$2a<p<3a.$$
This is exactly the required statement.