Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1970 IMO Problems/Problem 5"
Line 15: Line 15:
 
= CD^2,</math> so <math>CE^2 = CD^2 + DE^2</math>, so angle <math>CDE = 90^{\circ}</math>. But angle <math>CDB = 90^{\circ}</math>, so <math>CD</math> is
 
= CD^2,</math> so <math>CE^2 = CD^2 + DE^2</math>, so angle <math>CDE = 90^{\circ}</math>. But angle <math>CDB = 90^{\circ}</math>, so <math>CD</math> is
 
perpendicular to the plane <math>DAB</math>, and hence angle <math>CDA</math> = <math>90^{\circ}</math>. Similarly, angle <math>ADB = 90^{\circ}</math>.
 
perpendicular to the plane <math>DAB</math>, and hence angle <math>CDA</math> = <math>90^{\circ}</math>. Similarly, angle <math>ADB = 90^{\circ}</math>.
−
Hence <math>AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)</math>. But now we are done, because Cauchy's
+
Hence <math>AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)</math>.
−
inequality gives <math>(AB + BC + CA)^2 \le 3(AB^2 + BC^2 + CA^2)</math> (applied to vectors <math>(AB, BC, CA)</math>
+
 
−
and <math>(1, 1, 1))</math>. We have equality if and only if
+
But now we are done, because Cauchy's inequality (applied to vectors <math>(AB, BC, CA)</math> and <math>(1, 1, 1)</math>)
−
we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
+
gives <math>(AB + BC + CA)^2 \le 3(AB^2 + BC^2 + CA^2)</math>.
 +
 
 +
We have equality if and only if we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
  
  

Revision as of 16:38, 6 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 $AB = BC = CA.$


1970 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1970_IMO_Problems/Problem_5&oldid=236288"