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 | + | 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 ABC and let CH meet AB at X. Planes CED and ABC are perpendicular and AB is perpendicular to | + | 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 |
− | the line of intersection CE. Hence AB is perpendicular to the plane CDE and hence to ED. So 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. Also CB^2 = CE^2 + BE^2. Therefore CB^2 - BD^2 = CE^2 - DE^2. But 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, so CE^2 = CD^2 + DE^2, so angle CDE = | + | = 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 DAB, and hence angle CDA = | + | the plane <math>DAB</math>, and hence angle <math>CDA</math> = <math>90^{\circ}</math>. Similarly, angle <math>ADB = 90^{\circ}</math>. |
− | Hence AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2). 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 (AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2). We have equality if and only if | + | inequality gives <math>(AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2).</math> We have equality if and only if |
− | we have equality in Cauchy's inequality, which means AB = BC = CA. | + | we have equality in Cauchy's inequality, which means <math>AB = BC = CA.</math> |
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Revision as of 22:35, 18 July 2016
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 and are also right. Let be the intersection of the altitudes of and let meet at . Planes and are perpendicular and is perpendicular to the line of intersection . Hence is perpendicular to the plane and hence to . So Also Therefore But so , so angle . But angle , so is perpendicular to the plane , and hence angle = . Similarly, angle . Hence . But now we are done, because Cauchy's inequality gives We have equality if and only if we have equality in Cauchy's inequality, which means
1970 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |