Difference between revisions of "1984 AHSME Problems/Problem 27"
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<cmath>-2x^3 + 4 = 0</cmath> | <cmath>-2x^3 + 4 = 0</cmath> | ||
<cmath>-2(x^3 - 2) = 0; x^3 = 2</cmath> | <cmath>-2(x^3 - 2) = 0; x^3 = 2</cmath> | ||
− | Therefore, <math>AC^3 = 2 \boxed{\text{C}} </math>. | + | Therefore, <math>AC^3 = 2, \boxed{\text{C}} </math>. |
~elpianista227 | ~elpianista227 | ||
+ | |||
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=26|num-a=28}} | {{AHSME box|year=1984|num-b=26|num-a=28}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:44, 10 December 2024
Problem
In ,
is on
and
is on
. Also,
,
, and
. Find
.
Solution
Let and
. We have
by AA, so
. Substituting in known values gives
, so
. Also,
, and using the Pythagorean Theorem on
, we have
, so
. Using the Pythagorean Theorem on
gives
, or
. Now, we use the Pythagorean Theorem on
to get
. Substituting
into this gives
, or
. Simplifying this and moving all of the terms to one side gives
, and since
, we can divide by
to get
, from which we find that
.
Solution 2 (single-variable)
Let and draw
. Since
is the median of a right triangle, it follows that
. Then, since
and
are isoceles triangles, then they are considered similar. Therefore,
through similarity ratios. Finally, using the Pythagorean Theorem and the fact that
gives
and also
![]()
![]()
![]()
![]()
![]()
Therefore,
.
~elpianista227
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.