Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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<math> OE = 100\sqrt{2} = \frac{1}{2}OB</math>, <math>EF=\frac{1}{2}*BC=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}</math> | <math> OE = 100\sqrt{2} = \frac{1}{2}OB</math>, <math>EF=\frac{1}{2}*BC=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}</math> | ||
− | ===Solution 5=== | + | ===Solution 5 (Ptolemy's Theorem)=== |
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Revision as of 18:16, 22 February 2016
Contents
Problem
A quadrilateral is inscribed in a circle of radius . Three of the sides of this quadrilateral have length
. What is the length of the fourth side?
Solution
Solution 1 (Algebra)
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by for now, then multiply it back at the end of our solution.
Construct quadrilateral on the circle with
being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center
to
and
. Let the intersection of
and
be point
. Notice that
and
are perpendicular because
is a kite.
We set lengths equal to
. By the Pythagorean Theorem,
We solve for :
By Ptolemy's Theorem,
Substituting values,
Finally, we multiply back the that we divided by at the beginning of the problem to get
.
Solution 2 (Trigonometry Bash)
Construct quadrilateral on the circle with
being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center
to
and
. Apply law of cosines on
; let
. We get the following equation:
Substituting the values in, we get
Canceling out, we get
Because
,
, and
are congruent,
. To find the remaining side (
), we simply have to apply the law of cosines to
. Now, to find
, we can derive a formula that only uses
:
Plugging in
, we get
. Now, applying law of cosines on triangle
, we get
Solution 3 (Easier trig)
Construct quadrilateral on the circle
with
being the missing side. Then, drop perpendiculars from
and
to (extended) line
, and let these points be
and
, respectively. Also, let
. From Law of Cosines on
, we have
.
Now, since
is isosceles with
, we have that
. By SSS congruence, we have that
, so we have that
, so
.
Thus, we have
, so
. Similarly,
, and
.
Solution 4 (Just Geometry)
Label AD intercept OB at E and OC at F.
so,
we can get , similar
, we could get
,
, so
Solution 5 (Ptolemy's Theorem)
Let . Let
be the center of the circle. Then
is twice the altitude of
. Since
is isosceles we can compute its area to be
, hence
.
Now by Ptolemy's Theorem we have .
Solution 6
Using trig. Since all three sides equal , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths
by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is
. Similarly, the cosine is
.
Since there are three sides, and since
,we seek to find
.
First,
and
by Pythagorean.
See Also
2016 AMC 10A (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 10 Problems and Solutions |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.