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 1: Line 1:
 
==Problem==
 
==Problem==
 +
 
In the tetrahedron <math>ABCD</math>, angle <math>BDC</math> is a right angle.  Suppose that the foot <math>H</math> of the perpendicular from <math>D</math> to the plane <math>ABC</math> in the tetrahedron is the intersection of the altitudes of <math>\triangle ABC</math>.  Prove that
 
In the tetrahedron <math>ABCD</math>, angle <math>BDC</math> is a right angle.  Suppose that the foot <math>H</math> of the perpendicular from <math>D</math> to the plane <math>ABC</math> in the tetrahedron is the intersection of the altitudes of <math>\triangle ABC</math>.  Prove that
  
Line 7: Line 8:
  
 
For what tetrahedra does equality hold?
 
For what tetrahedra does equality hold?
 +
  
 
==Solution==
 
==Solution==
 +
 
Let us show first that angles <math>ADB</math> and <math>ADC</math> are also right. Let <math>H</math> be the intersection of the altitudes
 
Let us show first that angles <math>ADB</math> and <math>ADC</math> are also right. Let <math>H</math> be the intersection of the altitudes
 
of <math>ABC</math> and let <math>CH</math> meet <math>AB</math> at <math>E</math>. Planes <math>CED</math> and <math>ABC</math> are perpendicular and <math>AB</math> is perpendicular to
 
of <math>ABC</math> and let <math>CH</math> meet <math>AB</math> at <math>E</math>. Planes <math>CED</math> and <math>ABC</math> are perpendicular and <math>AB</math> is perpendicular to
Line 21: Line 24:
  
 
We have equality if and only if we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
 
We have equality if and only if we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
 +
 +
 +
==Solution 2==
 +
 +
Let <math>x = DH, a = BC, b = CA, c = AB</math>
 +
 +
[[File:Prob_1970_5.png|400px]]
 +
 +
The plan of this proof is to compute <math>HA, HB, HC</math> in terms of <math>a, b, c</math>, then
 +
compute <math>DA^2, DB^2, DC^2</math> in terms of <math>a, b, c, x</math>, impose the condition that
 +
<math>\angle BDC = \pi/2</math> to determine <math>x</math>, and calculate <math>DA^2 + DB^2 + DC^2</math> in
 +
terms of <math>a, b, c</math>.  The problem will become a simple inequality in <math>a, b, c</math>
 +
which will be easy to prove.
 +
 +
  
  

Revision as of 19:52, 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 $AB = BC = CA.$


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.



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=236298"