Difference between revisions of "1987 IMO Problems/Problem 2"
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+ | We are to prove that <math>[AKNM]=[ABC]</math> or equivilently, <math>[ABC]+[BNC]-[KNC]-[BMN]=[ABC]</math>. Thus, we are to prove that <math>[BNC]=[KNC]+[BMN]</math>. It is clear that since <math>\angle BAN=\angle NAC</math>, the segments <math>BN</math> and <math>NC</math> are equal. Thus, we have <math>[BNC]=\frac{1}{2}BN^2\sin BNC=\frac{1}{2}BN^2\sin A</math> since cyclic quadrilateral <math>ABNC</math> gives <math>\angle BNC=180-\angle A</math>. Thus, we are to prove that | ||
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+ | <math>\frac{1}{2}BN^2\sin A=[KNC]+[BMN]</math> | ||
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+ | <math>\Leftrightarrow \frac{1}{2}BN^2\sin A=\frac{1}{2}CN\cdot CK\sin NCA+\frac{1}{2}BN\cdot BM\sin NBA</math> | ||
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+ | <math>\Leftrightarrow BN\sin A=CK\sin NCA+BM\sin NBA</math> | ||
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+ | From the fact that <math>\angle BNC=180-\angle A</math> and that <math>BNC</math> is iscoceles, we find that <math>\angle NBC=\angle NCB=\frac{1}{2}A</math>. So, we have <math>BN\cos\frac{1}{2}A=\frac{1}{2}BC\Rightarrow BN=\frac{BC}{2\cos \frac{1}{2}A}</math>. So we are to prove that | ||
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+ | <math>\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA</math> | ||
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+ | <math>\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C\frac{1}{2}A)+BM\sin (C\frac{1}{2}A)</math> | ||
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+ | <math>\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)</math> | ||
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+ | We have <math>\sin C=\frac{KL}{CL}</math>,<math>\cos C=\frac{CK}{CL}</math>, <math>\cot\frac{1}{2}A=\frac{AK}{KL}=\frac{AM}{LM}</math>, <math>\sin B=\frac{LM}{BL}</math>,<math>\cos B=\frac{BM}{ML}</math>, and so we are to prove that | ||
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+ | <math>BC=CK(\frac{KL}{CL}\frac{AK}{KL})+\frac{CK}{CL})+BM(\frac{LM}{BL}\frac{AM}{LM}+\frac{BM}{ML})</math> | ||
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+ | <math>\Leftrightarrow BC=CK(\frac{AK}{CL}+\frac{CK}{CL})+BM(\frac{AM}{BL}+\frac{BM}{ML})</math> | ||
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+ | <math>\Leftrightarrow BC=\frac{CK\cdot AC}{CL}+\frac{BM\cdot AB}{BL}</math> | ||
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+ | <math>\Leftrightarrow BC=AC\cos C+AB\cos B</math> | ||
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+ | We shall show that this is true: Let the altitude from <math>A</math> touch <math>BC</math> at <math>A^\prime</math>. Then it is obvious that <math>AC\cos C=CA^\prime</math> and <math>AB\cos B=A^\prime B</math> and thus <math>AC\cos C+AB\cos B=BC</math>. | ||
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+ | Thus we have proven that <math>[AKNM]=[ABC]</math>. | ||
{{IMO box|num-b=1|num-a=3|year=1987}} | {{IMO box|num-b=1|num-a=3|year=1987}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 20:06, 11 September 2008
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 .
1987 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |