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Difference between revisions of "1970 IMO Problems/Problem 5"
Line 37: Line 37:
 
terms of <math>a, b, c</math>.  The problem will become a simple inequality in <math>a, b, c</math>
 
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.
 
which will be easy to prove.
 +
 +
Formulas for the distance from the orthocenter to the vertices are reasonably
 +
well known, but to make this solution self contained, we compute them here.
 +
From <math>\triangle ABE_B</math> we have that <math>BE_B = c \sin A</math>.  From <math>\triangle CBE</math>
 +
we have <math>BE = a \cos B</math>.  From <math>\triangle ABE_B \sim \triangle HBE</math> we have
 +
<math>\frac{HB}{c} = \frac{BE}{BE_B}</math>.  It follows that
 +
<math>HB = c \ \frac{a \cos B}{c \sin A} = \frac{a \cos B}{\sin A}</math>.
 +
 +
Similarly, we have <math>HC = \frac{a \cos C}{\sin A}</math> and
 +
<math>HA = \frac{b \cos A}{\sin B} = \frac{a \cos A}{\sin A}</math>.
 +
 +
(In the last equality we used the [[Law of Sines]]:
 +
<math>\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.</math>)
  
  

Revision as of 01:54, 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.

Formulas for the distance from the orthocenter to the vertices are reasonably well known, but to make this solution self contained, we compute them here. From $\triangle ABE_B$ we have that $BE_B = c \sin A$. From $\triangle CBE$ we have $BE = a \cos B$. From $\triangle ABE_B \sim \triangle HBE$ we have $\frac{HB}{c} = \frac{BE}{BE_B}$. It follows that $HB = c \ \frac{a \cos B}{c \sin A} = \frac{a \cos B}{\sin A}$.

Similarly, we have $HC = \frac{a \cos C}{\sin A}$ and $HA = \frac{b \cos A}{\sin B} = \frac{a \cos A}{\sin A}$.

(In the last equality we used the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$)



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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