Difference between revisions of "2013 Mock AIME I Problems/Problem 12"
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label("$C$", C, E); | label("$C$", C, E); | ||
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+ | ==See also== | ||
+ | * [[2013 Mock AIME I Problems/Problem 11|Preceded by Problem 11]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 13|Followed by Problem 13]] | ||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 21:37, 8 June 2022
Problem
In acute triangle , the orthocenter
lies on the line connecting the midpoint of segment
to the midpoint of segment
. If
, and the altitude from
has length
, find
.
Solution (easy coordinate bash)
Toss on the coordinate plane with ,
, and
, where
is a real number and
.
Then, the line connecting the midpoints of and
runs from
to
, or more simply the line
.
The orthocenter of will be at the intersection of the altitudes from
and
.
The slope of the altitude from is the negative reciprocal of the slope of
. The slope of
is
, and its negative reciprocal is
. Since the altitude from
passes through the origin, its equation is
.
The altitude from is the vertical line running through
which has equation
.
Thus the lines and
meet on the line
. Substituting the first equation into the second,
.
Multiplying both sides by , we have
.
This rearranges to the quadratic , and completing the square by adding
to each side gives us
. Thus
.
The cases where and
are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of
, so we consider the case where
.
So .
Thus
The cases where and
are shown below, labeled
and
, respectively, where the dotted line is a midline in both triangles. As you can see, the orthocenter falls perfectly on that line for both triangles, and the value of
is the same for both triangles.