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Difference between revisions of "1987 IMO Problems/Problem 2"
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<math>\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA</math>
 
<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\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)</math>
  
 
<math>\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)</math>
 
<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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Thus we have proven that <math>[AKNM]=[ABC]</math>.
 
Thus we have proven that <math>[AKNM]=[ABC]</math>.
  
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==Solution 2==
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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.
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 +
Hence, we have <math>[AKNM] = \frac{1}{2} AN \cdot KM = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM.</math> Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).
 +
 +
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.
 +
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==Solution 3==
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Proceed as in Solution 2. To prove that <math>\frac{KM}{AL} = \sin A</math>, consider that
 +
 +
<cmath>AL = \frac{AK}{\sin <KLA} = \frac{AK}{\sin <AKM} = \frac{KM}{\sin A}</cmath>
 +
 +
via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines.
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==See also==
 
{{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]]

Latest revision as of 14:39, 21 September 2014

Problem

In an acute-angled triangle $ABC$ the interior bisector of the angle $A$ intersects $BC$ at $L$ and intersects the circumcircle of $ABC$ again at $N$. From point $L$ perpendiculars are drawn to $AB$ and $AC$, the feet of these perpendiculars being $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

Solution

We are to prove that $[AKNM]=[ABC]$ or equivilently, $[ABC]+[BNC]-[KNC]-[BMN]=[ABC]$. Thus, we are to prove that $[BNC]=[KNC]+[BMN]$. It is clear that since $\angle BAN=\angle NAC$, the segments $BN$ and $NC$ are equal. Thus, we have $[BNC]=\frac{1}{2}BN^2\sin BNC=\frac{1}{2}BN^2\sin A$ since cyclic quadrilateral $ABNC$ gives $\angle BNC=180-\angle A$. Thus, we are to prove that

$\frac{1}{2}BN^2\sin A=[KNC]+[BMN]$

$\Leftrightarrow \frac{1}{2}BN^2\sin A=\frac{1}{2}CN\cdot CK\sin NCA+\frac{1}{2}BN\cdot BM\sin NBA$

$\Leftrightarrow BN\sin A=CK\sin NCA+BM\sin NBA$

From the fact that $\angle BNC=180-\angle A$ and that $BNC$ is iscoceles, we find that $\angle NBC=\angle NCB=\frac{1}{2}A$. So, we have $BN\cos\frac{1}{2}A=\frac{1}{2}BC\Rightarrow BN=\frac{BC}{2\cos \frac{1}{2}A}$. So we are to prove that

$\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA$

$\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)$

$\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)$

We have $\sin C=\frac{KL}{CL}$,$\cos C=\frac{CK}{CL}$, $\cot\frac{1}{2}A=\frac{AK}{KL}=\frac{AM}{LM}$, $\sin B=\frac{LM}{BL}$,$\cos B=\frac{BM}{ML}$, and so we are to prove that

$BC=CK(\frac{KL}{CL}\frac{AK}{KL})+\frac{CK}{CL})+BM(\frac{LM}{BL}\frac{AM}{LM}+\frac{BM}{ML})$

$\Leftrightarrow BC=CK(\frac{AK}{CL}+\frac{CK}{CL})+BM(\frac{AM}{BL}+\frac{BM}{ML})$

$\Leftrightarrow BC=\frac{CK\cdot AC}{CL}+\frac{BM\cdot AB}{BL}$

$\Leftrightarrow BC=AC\cos C+AB\cos B$

We shall show that this is true: Let the altitude from $A$ touch $BC$ at $A^\prime$. Then it is obvious that $AC\cos C=CA^\prime$ and $AB\cos B=A^\prime B$ and thus $AC\cos C+AB\cos B=BC$.

Thus we have proven that $[AKNM]=[ABC]$.

Solution 2

Clearly, $AKLM$ is a kite, so its diagonals are perpendicular. Furthermore, we have triangles $ABN$ and $ALC$ similar because two corresponding angles are equal.

Hence, we have $[AKNM] = \frac{1}{2} AN \cdot KM = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM.$ Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).

But in (right) triangle $AKL$, we have $<LAB = <A/2$. Furthermore, if $Q$ is the intersection of diagonals $AL$ and $KM$ we have $Q$ the midpoint of $KM$ and $KQ$ an altitude of $AKL$, so \[\frac{KM}{2} = \frac{AK \cdot KL}{AL} = \frac{AL \sin <A/2 \cdot AL \cos <A/2}{AL} = \frac{AL \sin <A}{2},\] so $\frac{KM}{AL} = \sin <A$. Hence \[[AKNM] = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM = \frac{1}{2} AB \cdot AC \sin <A = [ABC],\] as desired.

Solution 3

Proceed as in Solution 2. To prove that $\frac{KM}{AL} = \sin A$, consider that

\[AL = \frac{AK}{\sin <KLA} = \frac{AK}{\sin <AKM} = \frac{KM}{\sin A}\]

via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines.

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
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