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Difference between revisions of "1970 IMO Problems/Problem 5"
 
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==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
  
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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
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= 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>
 +
 
 +
 
 +
==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.
 +
 
 +
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>  We want to have
 +
<math>\sin A</math> at the denominator to simplify computations.)
 +
 
 +
Then <math>DA^2 = x^2 + \frac{a^2 \cos^2 A}{\sin^2 A},
 +
DB^2 = x^2 + \frac{a^2 \cos^2 B}{\sin^2 A},
 +
DC^2 = x^2 + \frac{a^2 \cos^2 C}{\sin^2 A}</math>.
 +
 
 +
Since <math>\angle BDC = \pi/2, DB^2 + DC^2 = a^2</math>.  This yields
 +
<math>x^2 = \frac{a^2}{2 \sin^2 A} \ (\sin^2 A - \cos^2 B - \cos^2 C)</math>.
 +
 
 +
Using this value for <math>x^2</math> we get
 +
<math>DA^2 + DB^2 + DC^2 =
 +
\frac{a^2}{2 \sin^2 A} \ (3 \sin^2 A + 2 \cos^2 A - \cos^2 B - \cos^2 C)</math>
 +
 
 +
After some simplifications, and using again the [[Law of Sines]], this becomes
 +
 
 +
<math>DA^2 + DB^2 + DC^2 =
 +
\frac{a^2}{2 \sin^2 A} \ (\sin^2 A + \sin^2 B + \sin^2 C) =
 +
\frac{a^2}{2} + \frac{a^2}{2 \sin^2 A} \sin^2 B + \frac{a^2}{2 \sin^2 A} \sin^2 C =</math>
 +
 
 +
<math>\frac{a^2}{2} + \frac{b^2}{2 \sin^2 B} \sin^2 B + \frac{c^2}{2 \sin^2 C} \sin^2 C =
 +
\frac{1}{2} \ (a^2 + b^2 + c^2)</math>.
 +
 
 +
What we need to prove then is that <math>(a + b + c)^2 \le 3(a^2 + b^2 + c^2)</math>.  We could
 +
invoke the [[Cauchy-Schwarz Inequality]] like in the first solution, but we can
 +
just as well make the computations, and notice that this is equivalent to
 +
 
 +
<math>2 a^2 + 2 b^2 + 2 c^2 -2ab - 2bc - 2ca \ge 0</math> or
 +
<math>(a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0</math>, which is always true.
 +
 
 +
We also see that we have equality if and only if <math>a = b = c</math>.
 +
 
 +
[Solution by pf02, December 2024]
  
  

Latest revision as of 02:40, 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: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$ We want to have $\sin A$ at the denominator to simplify computations.)

Then $DA^2 = x^2 + \frac{a^2 \cos^2 A}{\sin^2 A}, DB^2 = x^2 + \frac{a^2 \cos^2 B}{\sin^2 A}, DC^2 = x^2 + \frac{a^2 \cos^2 C}{\sin^2 A}$.

Since $\angle BDC = \pi/2, DB^2 + DC^2 = a^2$. This yields $x^2 = \frac{a^2}{2 \sin^2 A} \ (\sin^2 A - \cos^2 B - \cos^2 C)$.

Using this value for $x^2$ we get $DA^2 + DB^2 + DC^2 = \frac{a^2}{2 \sin^2 A} \ (3 \sin^2 A + 2 \cos^2 A - \cos^2 B - \cos^2 C)$

After some simplifications, and using again the Law of Sines, this becomes

$DA^2 + DB^2 + DC^2 = \frac{a^2}{2 \sin^2 A} \ (\sin^2 A + \sin^2 B + \sin^2 C) = \frac{a^2}{2} + \frac{a^2}{2 \sin^2 A} \sin^2 B + \frac{a^2}{2 \sin^2 A} \sin^2 C =$

$\frac{a^2}{2} + \frac{b^2}{2 \sin^2 B} \sin^2 B + \frac{c^2}{2 \sin^2 C} \sin^2 C = \frac{1}{2} \ (a^2 + b^2 + c^2)$.

What we need to prove then is that $(a + b + c)^2 \le 3(a^2 + b^2 + c^2)$. We could invoke the Cauchy-Schwarz Inequality like in the first solution, but we can just as well make the computations, and notice that this is equivalent to

$2 a^2 + 2 b^2 + 2 c^2 -2ab - 2bc - 2ca \ge 0$ or $(a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0$, which is always true.

We also see that we have equality if and only if $a = b = c$.

[Solution by pf02, December 2024]


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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