Difference between revisions of "2002 AMC 12P Problems/Problem 24"
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Consider triangles <math>EPS</math>, <math>EQT</math>, and <math>ERU</math>. Each of these triangles have a right angle and an angle equal to the dihedral angle of the tetrahedron, so they are all similar by AA similarity. In particular, we know that <math>\frac{EP}{ES} = \frac{EQ}{ET} = \frac{ER}{EU} = \frac{EP+EQ+ER}{ES+ET+EU} = \frac{s}{S}</math>. | Consider triangles <math>EPS</math>, <math>EQT</math>, and <math>ERU</math>. Each of these triangles have a right angle and an angle equal to the dihedral angle of the tetrahedron, so they are all similar by AA similarity. In particular, we know that <math>\frac{EP}{ES} = \frac{EQ}{ET} = \frac{ER}{EU} = \frac{EP+EQ+ER}{ES+ET+EU} = \frac{s}{S}</math>. | ||
− | It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\ | + | It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\sin(\angle DSE)</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=23|num-a=25}} | {{AMC12 box|year=2002|ab=P|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:53, 10 March 2024
Problem
Let be a regular tetrahedron and Let be a point inside the face Denote by the sum of the distances from to the faces and by the sum of the distances from to the edges Then equals
Solution
Assume points , , and are on faces , , and respectively such that , , and .
Assume points , , and are on edges , , and respectively such that , , and .
Consider triangles , , and . Each of these triangles have a right angle and an angle equal to the dihedral angle of the tetrahedron, so they are all similar by AA similarity. In particular, we know that .
It remains to find , or equivalently, .
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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