Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath> | Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath> | ||
+ | |||
+ | ==Solution 5== | ||
+ | <center><img>https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZi84L2E4MzA5YjkzYjhmMmMxYmY0YTVjMTNjOTJmNDVkZjE5MWMwZDZjLnBuZw==&rn=U2NyZWVuIFNob3QgMjAyMS0wOC0wNiBhdCA3LjMwLjEwIFBNLnBuZw==</img> | ||
+ | </center> | ||
+ | |||
+ | Since <math>ED</math> is a midline of <math>\triangle CAB,</math> we have that <math>\triangle CED \sim \triangle CAB</math> with a side length ratio of <math>1:2.</math> | ||
+ | |||
+ | Consider a homothety of scale factor <math>2</math> with on <math>\triangle CED</math> with respect to point <math>C.</math> Note that this sends <math>(CEDX)</math> to <math>(ABCC')</math> with <math>CX=XC'.</math> By properties of homtheties, <math>C,X,</math> and <math>C'</math> are collinear. Similarly, we obtain that <math>BX=XB',</math> with all three points collinear. Let <math>O</math> denote the circumcenter of <math>\triangle ABC.</math> It is well-known that <math>OX \perp CC'</math> and analogously <math>OX \perp BB'.</math> However, there is only one perpendicular line to <math>OX</math> passing through <math>X,</math>, therefore, <math>O</math> coincides with <math>X.</math> | ||
+ | |||
+ | It follows that <math>AX=BX=CX=R,</math> where <math>R</math> is the circumradius of <math>\triangle ABC,</math> and this can be computed using the formula <cmath>R=\frac{abc}{4[ABC]},</cmath> from which we quickly obtain <cmath>R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.</cmath> | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:31, 6 August 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
Solution 5
Since is a midline of
we have that
with a side length ratio of
Consider a homothety of scale factor with on
with respect to point
Note that this sends
to
with
By properties of homtheties,
and
are collinear. Similarly, we obtain that
with all three points collinear. Let
denote the circumcenter of
It is well-known that
and analogously
However, there is only one perpendicular line to
passing through
, therefore,
coincides with
It follows that where
is the circumradius of
and this can be computed using the formula
from which we quickly obtain
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 |
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