Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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After a few algebraic manipulations: | After a few algebraic manipulations: | ||
− | <math>\Rightarrow R=\frac{65}{8} \Rightarrow | + | <math>\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}</math>. |
===Solution 3 (Homothety)=== | ===Solution 3 (Homothety)=== |
Revision as of 17:59, 14 July 2021
Contents
Problem
Triangle has
, and
. The points
, and
are the midpoints of
, and
respectively. Let
be the intersection of the circumcircles of
and
. What is
?
Solutions
Solution 1 (Coordinates)
Let us also consider the circumcircle of .
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of which is
, Also, since
.
is cyclic, similarly,
and
are also cyclic. With this, we know that the circumcircles of
,
and
all intersect at
, so
is
.
The question now becomes calculate the sum of distance from each vertices to the circumcenter.
We can calculate the distances with coordinate geometry. (Note that because
is the circumcenter.)
Let ,
,
,
Then is on the line
and also the line with slope
that passes through
.
So
and
Solution 2 (Algebra)
Consider an additional circumcircle on . After drawing the diagram, it is noticed that each triangle has side values:
,
,
. Thus they are congruent, and their respective circumcircles are. By inspection, we see that
,
, and
are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of
. We can find the circumradius quite easily with the formula
, such that
and
is the circumradius. Since
:
After a few algebraic manipulations:
.
Solution 3 (Homothety)
Let be the circumcenter of
and
denote the length of the altitude from
Note that a homothety centered at
with ratio
takes the circumcircle of
to the circumcircle of
. It also takes the point diametrically opposite
on the circumcircle of
to
Therefore,
lies on the circumcircle of
Similarly, it lies on the circumcircle of
By Pythagorean triples,
Finally, our answer is
Solution 4 (basically Solution 1 but without coordinates)
Since Solution 1 has already proven that the circumcenter of coincides with
, we'll go from there. Note that the radius of the circumcenter of any given triangle is
, and since
and
, it can be easily seen that
and therefore our answer is
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.