1973 USAMO Problems/Problem 1
Contents
Problem
Two points 
and 
lie in the interior of a regular tetrahedron 
. Prove that angle 
.
Solutions
Solution 1
Let the side length of the regular tetrahedron be 
. Link and extend 
to meet the plane containing triangle 
at 
; link 
and extend it to meet the same plane at 
. We know that and are inside triangle and that 
Now let’s look at the plane containing triangle with points and inside the triangle. Link and extend 
on both sides to meet the sides of the triangle at 
and 
, on 
and on 
. We have 
But since and are interior of the tetrahedron, points and cannot be both at the vertices and 
, 
. Therefore, 
.
Solution with graphs posted at
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1
Solution 2 (similar to solution 1, but full proof)
Let the side length of the regular tetrahedron be 
.
Extendand
to intersect the plane
at points
and
inside
, respectively. We can observe that
.
and 
inside 
, respectively. We can observe that 
.
Extendto intersect the sides of
and
. We can observe that
. We assume that, WLOG,
and
.
to intersect the sides of 
and 
. We can observe that 
. We assume that, WLOG, 
and Let the ratiosand
. We can see that
. It follows from Stewart's theorem that \begin{eqnarray*} AP^2=AD^2\mu+AB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s\\ AQ^2=AC^2\lambda+AB^2(1-\lambda)-\lambda(1-\lambda)BC^2=(\lambda^2-\lambda^2+1)s \end{eqnarray*}.
and 
. We can see that 
. It follows from Stewart's theorem that
\begin{eqnarray*}
AP^2=AD^2\mu+AB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s\\
AQ^2=AC^2\lambda+AB^2(1-\lambda)-\lambda(1-\lambda)BC^2=(\lambda^2-\lambda^2+1)s
\end{eqnarray*}.
Furthermore, we have
\[
PC^2=CD^2\mu+CB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s
\]
Using this, we can conclude that
\begin{eqnarray*}
PQ^2&&=PC^2\lambda+(BD\mu)^2(1-\lambda)-\lambda(1-\lambda)BC^2\\
&&=(\mu\lambda-\mu-\lambda+2)s^2
\end{eqnarray*}.
By cosine law, we have
\begin{eqnarray*}
\cos \angle PAQ&&=\frac{PA^2+QA^2-PQ^2}{2PA\cdot QA}\\
&&=\frac{\mu\lambda-\mu-\lambda+2}{2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}}
\end{eqnarray*}
for 
.
.
We first find a lower bound of the numerator. We writeas a linear function of
by letting
be constant. Since
, the coefficient
of
. Hence,
. Now we find an upper bound of the denominator. We complete the square in the expression
which is least when
.
as a linear function of 
by letting 
be constant. Since 
, the coefficient 
of 
. Hence, 
.
Now we find an upper bound of the denominator. We complete the square in the expression 
which is least when 
.
Hence,, implying that $\angle P''AQ''<60\degree$ (Error compiling LaTeX. Unknown error_msg).
, implying that Therefore, $\angle PAQ=\angle P'AQ'<\angle P''AQ''<60\degree$ (Error compiling LaTeX. Unknown error_msg).
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
hurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification.
See also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
