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Difference between revisions of "1970 IMO Problems/Problem 5"

m (→‎Solution)
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==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>X</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
 
the line of intersection <math>CE</math>. Hence <math>AB</math> is perpendicular to the plane <math>CDE</math> and hence to <math>ED</math>. So <math>BD^2 =
 
the line of intersection <math>CE</math>. Hence <math>AB</math> is perpendicular to the plane <math>CDE</math> and hence to <math>ED</math>. So <math>BD^2 =
 
DE^2 + BE^2.</math> Also <math>CB^2 = CE^2 + BE^2.</math> Therefore <math>CB^2 - BD^2 = CE^2 - DE^2.</math> But <math>CB^2 - BD^2
 
DE^2 + BE^2.</math> Also <math>CB^2 = CE^2 + BE^2.</math> Therefore <math>CB^2 - BD^2 = CE^2 - DE^2.</math> But <math>CB^2 - BD^2
−
= 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
+
= 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
−
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>. But now we are done, because Cauchy's
−
inequality gives <math>(AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2).</math> We have equality if and only if
+
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
 
we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
 
we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math>
  

Revision as of 16:33, 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 gives $(AB + BC + CA)^2 \le 3(AB^2 + BC^2 + CA^2)$ (applied to vectors $(AB, BC, CA)$ and $(1, 1, 1))$. 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
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