Difference between revisions of "1972 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
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| + | ===Solution 1=== | ||
Proof by contradiction: Obviously, triangles ACD, DCB, and ABC are all congruent (the three sides are all equal). For the sake of clarity, triangle ABC is the "base", and D is the "pointed" vertex. Assume <DCB is obtuse (this proof will work for any other angle as well). Therefore angles <ABD and <DCA are also obtuse because of the congruence. However, this will mean that AB and AC will be extended outwards, and as these angles approach over 90 degrees, CA and AB will be parallel at 90, and then will extend outwards, making the point A non-existent, which obviously is a contradiction of the fact that ABCD is a tetrahedron. | Proof by contradiction: Obviously, triangles ACD, DCB, and ABC are all congruent (the three sides are all equal). For the sake of clarity, triangle ABC is the "base", and D is the "pointed" vertex. Assume <DCB is obtuse (this proof will work for any other angle as well). Therefore angles <ABD and <DCA are also obtuse because of the congruence. However, this will mean that AB and AC will be extended outwards, and as these angles approach over 90 degrees, CA and AB will be parallel at 90, and then will extend outwards, making the point A non-existent, which obviously is a contradiction of the fact that ABCD is a tetrahedron. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | |||
| + | Suppose <math>\triangle ABD</math> is fixed. | ||
| + | By the equality conditions, it follows that the maximal possible value of <math>BC</math> occurs when the four vertices are coplanar, with <math>C</math> on the opposite side of <math>\overline{AD}</math> as <math>B</math>. | ||
| + | In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable. | ||
| + | |||
| + | For the sake of contradiction, suppose <math>\angle ABD</math> is non-acute. | ||
| + | Then, <math>(AD)^2\geq (AB)^2+(BD)^2</math>. | ||
| + | In our optimal case noted above, <math>ACDB</math> is a parallelogram, so | ||
| + | <cmath>\begin{align*} | ||
| + | 2(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\ | ||
| + | &= 2(AD)^2 \\ | ||
| + | &\geq 2(BD)^2+2(AB)^2. | ||
| + | \end{align*}</cmath> | ||
| + | However, as stated, equality cannot be attained, so we get our desired contradiction. | ||
==See also== | ==See also== | ||
Revision as of 19:13, 2 April 2010
Problem
A given tetrahedron 
is isosceles, that is, 
. Show that the faces of the tetrahedron are acute-angled triangles.
Solution
Solution 1
Proof by contradiction: Obviously, triangles ACD, DCB, and ABC are all congruent (the three sides are all equal). For the sake of clarity, triangle ABC is the "base", and D is the "pointed" vertex. Assume <DCB is obtuse (this proof will work for any other angle as well). Therefore angles <ABD and <DCA are also obtuse because of the congruence. However, this will mean that AB and AC will be extended outwards, and as these angles approach over 90 degrees, CA and AB will be parallel at 90, and then will extend outwards, making the point A non-existent, which obviously is a contradiction of the fact that ABCD is a tetrahedron.
Solution 2
Suppose 
is fixed.
By the equality conditions, it follows that the maximal possible value of 
occurs when the four vertices are coplanar, with 
on the opposite side of 
as 
.
In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
For the sake of contradiction, suppose 
is non-acute.
Then, 
.
In our optimal case noted above, 
is a parallelogram, so
However, as stated, equality cannot be attained, so we get our desired contradiction.
See also
| 1972 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
