Difference between revisions of "2014 USAJMO Problems/Problem 2"
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The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | ||
− | Note: It's easy to show that for any point <math>H</math> on <math>\verline{PQ}</math>, Points B and C can be validly defined to make an acute, non-equilateral triangle | + | Note: It's easy to show that for any point <math>H</math> on <math>\verline{PQ}</math> except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle. |
Revision as of 02:54, 10 May 2014
Problem
Let be a non-equilateral, acute triangle with
, and let
and
denote the circumcenter and orthocenter of
, respectively.
(a) Prove that line intersects both segments
and
.
(b) Line intersects segments
and
at
and
, respectively. Denote by
and
the respective areas of triangle
and quadrilateral
. Determine the range of possible values for
.
Solution
Claim: is the reflection of
over the angle bisector of
(henceforth 'the' reflection)
Proof: Let be the reflection of
, and let
be the reflection of
.
Then reflection takes to
.
is equilateral, and
lies on the perpendicular bisector of
It's well known that lies strictly inside
, meaning that
from which it follows that $\overline{BH'} \perp \verline{AC}$ (Error compiling LaTeX. Unknown error_msg) . Similarly, $\overline{CH'} \perp \verline{AB}$ (Error compiling LaTeX. Unknown error_msg). Since
lies on two altitudes,
is the orthocenter, as desired.
So is perpendicular to the angle bisector of
, which is the same line as the angle bisector of
, meaning that
is equilateral.
Let its side length be , and let
, where
because O lies strictly within
, as must
, the reflection of
. Also, it's easy to show that if
in a general triangle, it's equlateral, and we know that
is not equilateral. Hence
. Let
intersect
at
.
Since and
are 30-60-90 triangles,
Similarly,
The ratio is
The denominator equals
where
can equal any value in
except
. Therefore, the denominator can equal any value in
, and the ratio is any value in
Note: It's easy to show that for any point on $\verline{PQ}$ (Error compiling LaTeX. Unknown error_msg) except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.