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==Video Solution== | ==Video Solution== |
Revision as of 13:12, 22 September 2023
Contents
- 1 Problem
- 2 Solution 1 (Complex Numbers)
- 3 Solution 2 (Trigonometry)
- 4 Solution 3 (Complex Numbers and Trigonometry)
- 5 Solution 4 (Trigonometry)
- 6 Solution 5 (Law of Cosines)
- 7 Solution 6 (Ruler Measure)
- 8 Solution 7 (Pythagorean Theorem and Trig)
- 9 Solution 8 (Roots of Unity)
- 10 Video Solution
- 11 See Also
Problem
The figure below depicts a regular -gon inscribed in a unit circle.
What is the sum of the
th powers of the lengths of all
of its edges and diagonals?
Solution 1 (Complex Numbers)
There are segments whose lengths are
,
segments whose lengths are
,
segments whose lengths are
.
Therefore, the sum of the th powers of these lengths is
where the fourth from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Trigonometry)
There are segments whose lengths are
,
segments whose lengths are
,
segments whose lengths are
.
Therefore, the sum of the th powers of these lengths is
where the second from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Complex Numbers and Trigonometry)
As explained in Solutions 1 and 2, what we are trying to find is . Using trig we get
Like in the second solution, we also use the fact that
, which admittedly might need some explanation. Notice that
In the brackets we have the sum of the roots of the polynomial
. These sum to
by Vieta’s formulas, and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question:
~obscene_kangaroo
Solution 4 (Trigonometry)
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
We end up with
Using the quadratic formula, we find the solutions for
to be
and
. Because
is impossible,
.
With this result, following similar to steps to Solutions 2 and 3 will get
~lordf
Solution 5 (Law of Cosines)
Let ,
, and
be the lengths of the chords with arcs
,
and
respectively.
Then by the law of cosines we get:
The answer is then just
(since there's
of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by
.
(Use the identity that
.)
- SAHANWIJETUNGA
Solution 6 (Ruler Measure)
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by
.
Measuring the sides of the circle gets cm. The shorter diagonals are
cm, and the longest diagonals measure
cm. Thus, we'd like to estimate
We know is slightly less than
Let's approximate it as 1 for now. Thus,
Next, is slightly more than
We know
slightly more than
so we can approximate
as
Thus,
Finally, is slightly less than
We say it's around
so then
Adding what we have, we get as our estimate. We see
is very close to our estimate, so we circle it and are happy that we successfully cheesed an AMC 12B problem 24.
~sirswagger21
Solution 7 (Pythagorean Theorem and Trig)
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle which are
for integer
Then, we notice there are three types of diagonals: the ones with chords of arcs
and
We notive there are here are
of each type of diagonal. Then, we use the pythagorean theorem to find the distance from
to
:
By the cosine double angle identity,
This means that
Substituting this in,
Summing this up for
(These equalities are based on
and
)
Finally, because there are
of each type of diagonal, the answer is
Solution 8 (Roots of Unity)
for each vertex so we need to multiply by . The formula is
-maxamc
Video Solution
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Label the vertices A, B, C ... F ;
Let AB = BC = ... = FG = GA = x ;
Let AC = BD = ... = FA = GB = y ;
Let AD = BE = ... = FB = GC = z ;
Thus, we have to find 7(x^4 + y^4 + z^4)
Using Ptolemy's theorem on various cyclic quadrilaterals, we find:
xy + y^2 = z^2 ;
yz + x^2 = z^2 ;
xz + xy = yz ;
x^2 + xz = y^2
- alphamugamma
See Also
2022 AMC 12B (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.