Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 5"
(Created page with "==Problem== In triangle <math>ABC,</math> <math>AB=36,BC=40,CA=44.</math> The bisector of angle <math>A</math> meet <math>BC</math> at <math>D</math> and the circumcircle at <ma...") |
(→Solution) |
||
Line 7: | Line 7: | ||
− | Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math> , and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>. Thus, <math>DH=CD-CH=22-20=2</math>. | + | Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math>, and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>. Thus, <math>DH=CD-CH=22-20=2</math>. |
Revision as of 21:55, 1 January 2012
Problem
In triangle
The bisector of angle
meet
at
and the circumcircle at
different from
. Calculate the value of
Solution
because they are both subscribed by arc
.
because they are both subscribed by arc
. Hence
, because
. Then
is isosceles.
Let be the foot of the perpendicular from
to
. As
is isosceles, it follows that
is the midpoint of
, and so
. From the angle bisector theorem,
. We have
. Solving this system of equations yields
. Thus,
.
because they are vertical angles. It was shown
, and so
by
similarity. Then
and so
.
Then by the Pythagorean Theorem on ,
. Also from
,
. Subtracting these equations yields
, and so
.