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Difference between revisions of "1970 IMO Problems/Problem 5"
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TO BE CONTINUED.
  
  

Revision as of 01:11, 7 December 2024

Problem

In the tetrahedron $ABCD$, angle $BDC$ is a right angle. Suppose that the foot $H$ of the perpendicular from $D$ to the plane $ABC$ in the tetrahedron is the intersection of the altitudes of $\triangle ABC$. Prove that

$( AB+BC+CA )^2 \leq 6( AD^2 + BD^2 + CD^2 )$.

For what tetrahedra does equality hold?


Solution

Let us show first that angles $ADB$ and $ADC$ are also right. Let $H$ be the intersection of the altitudes of $ABC$ and let $CH$ meet $AB$ at $E$. Planes $CED$ and $ABC$ are perpendicular and $AB$ is perpendicular to the line of intersection $CE$. Hence $AB$ is perpendicular to the plane $CDE$ and hence to $ED$. So $BD^2 = DE^2 + BE^2.$ Also $CB^2 = CE^2 + BE^2.$ Therefore $CB^2 - BD^2 = CE^2 - DE^2.$ But $CB^2 - BD^2 = CD^2,$ so $CE^2 = CD^2 + DE^2$, so angle $CDE = 90^{\circ}$. But angle $CDB = 90^{\circ}$, so $CD$ is perpendicular to the plane $DAB$, and hence angle $CDA$ = $90^{\circ}$. Similarly, angle $ADB = 90^{\circ}$. Hence $AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)$.

But now we are done, because Cauchy's inequality (applied to vectors $(AB, BC, CA)$ and $(1, 1, 1)$) gives $(AB + BC + CA)^2 \le 3(AB^2 + BC^2 + CA^2)$.

We have equality if and only if we have equality in Cauchy's inequality, which means $AB = BC = CA.$


Solution 2

Let $x = DH, a = BC, b = CA, c = AB$

Prob 1970 5.png

The plan of this proof is to compute $HA, HB, HC$ in terms of $a, b, c$, then compute $DA^2, DB^2, DC^2$ in terms of $a, b, c, x$, impose the condition that $\angle BDC = \pi/2$ to determine $x$, and calculate $DA^2 + DB^2 + DC^2$ in terms of $a, b, c$. The problem will become a simple inequality in $a, b, c$ which will be easy to prove.



TO BE CONTINUED.


1970 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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