Difference between revisions of "1987 IMO Problems/Problem 2"
Archimedes1 (talk | contribs) m (→Solution) |
(→Solution) |
||
Line 33: | Line 33: | ||
Thus we have proven that <math>[AKNM]=[ABC]</math>. | Thus we have proven that <math>[AKNM]=[ABC]</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Clearly, <math>AKLM</math> is a kite, so its diagonals are perpendicular. Furthermore, we have triangles <math>ABN</math> and <math>ALC</math> similar because two corresponding angles are equal. | ||
+ | |||
+ | Hence, we have <math>[AKNM] = \frac{1}{2} AN \cdot KM = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM.</math> | ||
+ | |||
+ | But in (right) triangle <math>AKL</math>, we have <math><LAB = <A/2</math>. Furthermore, if <math>Q</math> is the intersection of diagonals <math>AL</math> and <math>KM</math> we have <math>Q</math> the midpoint of <math>KM</math> and <math>KQ</math> an altitude of <math>AKL</math>, so | ||
+ | <cmath>\frac{KM}{2} = \frac{AK \cdot KL}{AL} = \frac{AL \sin <A/2 \cdot AL \cos <A/2}{AL} = \frac{AL \sin <A}{2},</cmath> | ||
+ | so <math>\frac{KM}{AL} = \sin <A</math>. Hence <cmath>[AKNM] = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM = \frac{1}{2} AB \cdot AC \sin <A = [ABC],</cmath> as desired. | ||
==See also== | ==See also== |
Revision as of 14:33, 21 September 2014
Contents
[hide]Problem
In an acute-angled triangle the interior bisector of the angle intersects at and intersects the circumcircle of again at . From point perpendiculars are drawn to and , the feet of these perpendiculars being and respectively. Prove that the quadrilateral and the triangle have equal areas.
Solution
We are to prove that or equivilently, . Thus, we are to prove that . It is clear that since , the segments and are equal. Thus, we have since cyclic quadrilateral gives . Thus, we are to prove that
From the fact that and that is iscoceles, we find that . So, we have . So we are to prove that
We have ,, , ,, and so we are to prove that
We shall show that this is true: Let the altitude from touch at . Then it is obvious that and and thus .
Thus we have proven that .
Solution 2
Clearly, is a kite, so its diagonals are perpendicular. Furthermore, we have triangles and similar because two corresponding angles are equal.
Hence, we have
But in (right) triangle , we have . Furthermore, if is the intersection of diagonals and we have the midpoint of and an altitude of , so so . Hence as desired.
See also
1987 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |