Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 14"
Noobmaster m (talk | contribs) (→Solution) |
Noobmaster m (talk | contribs) (→Solution) |
||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | Notice that <math>X</math> is the reflection of <math>H</math> through the midpoint of <math>BC</math>. So by reflecting the orthocenter lemma we know that <math>AX</math> is a diametre of <math>(ABC)</math>. [<math>(ABC)</math> means circumcircle of <math>\triangle{ABC}</math>. Simillarly <math>BY</math> and <math>CZ</math> also diametre of <math>(ABC)</math> | + | Notice that <math>X</math> is the reflection of <math>H</math> through the midpoint of <math>BC</math>. So by reflecting the orthocenter lemma we know that <math>AX</math> is a diametre of <math>(ABC)</math>. [<math>(ABC)</math> means circumcircle of <math>\triangle{ABC}</math>]. Simillarly <math>BY</math> and <math>CZ</math> also diametre of <math>(ABC)</math>. So we need to find <math>6R</math> where <math>R</math> is the radius of <math>(ABC)</math> |
Now by cosine rule we get, | Now by cosine rule we get, |
Revision as of 10:04, 9 August 2022
Contents
Problem
Let be a triangle such that
,
, and
. Let
be the orthocenter of
(intersection of the altitudes). Let
be the midpoint of
,
be the midpoint of
, and
be the midpoint of
. Points
,
, and
are constructed on
,
, and
, respectively, such that
is the midpoint of
,
is the midpoint of
, and
is the midpoint of
. Find
.
Solution
Notice that is the reflection of
through the midpoint of
. So by reflecting the orthocenter lemma we know that
is a diametre of
. [
means circumcircle of
]. Simillarly
and
also diametre of
. So we need to find
where
is the radius of
Now by cosine rule we get,
So
Now by sine rule we get,
So required answer is
By NOOBMASTER_M
Solution
See also
Mock AIME 5 2005-2006 (Problems, Source) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |