Difference between revisions of "1970 IMO Problems/Problem 5"
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==Solution== | ==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 X. 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°. But angle CDB = 90°, so CD is perpendicular to | ||
| + | the plane DAB, and hence angle CDA = 90°. Similarly, angle ADB = 90°. | ||
| + | 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 = 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. | ||
| + | |||
{{IMO box|year=1970|num-b=4|num-a=6}} | {{IMO box|year=1970|num-b=4|num-a=6}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 07:06, 1 August 2009
Problem
In the tetrahedron 
, angle 
is a right angle. Suppose that the foot 
of the perpendicular from 
to the plane 
in the tetrahedron is the intersection of the altitudes of 
. Prove that
.
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 X. 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°. But angle CDB = 90°, so CD is perpendicular to the plane DAB, and hence angle CDA = 90°. Similarly, angle ADB = 90°. 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 = 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 | ||
