Difference between revisions of "2002 AMC 12P Problems/Problem 24"
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It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\sin(\angle DSE)</math>. | It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\sin(\angle DSE)</math>. | ||
− | + | We know <math>SE = \frac{1}{3}DS</math> by the centroid property. Therefore, <math>\cos(\angle DSE) = \frac{1}{3}</math>, so <math>\sin(\angle DSE) = \sqrt{1-(\frac{1}{3})^2} = \boxed {\text{(B) }\frac{2 \sqrt{2}}{3}}</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}} |
Latest revision as of 19:04, 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,
.
We know by the centroid property. Therefore,
, so
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.