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Difference between revisions of "1988 IMO Problems/Problem 5"
 
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==Problem==
 
==Problem==
 
In a right-angled triangle <math> ABC</math> let <math> AD</math> be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles <math> ABD, ACD</math> intersect the sides <math> AB, AC</math> at the points <math> K,L</math> respectively. If <math> E</math> and <math> E_1</math> dnote the areas of triangles <math> ABC</math> and <math> AKL</math> respectively, show that
 
In a right-angled triangle <math> ABC</math> let <math> AD</math> be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles <math> ABD, ACD</math> intersect the sides <math> AB, AC</math> at the points <math> K,L</math> respectively. If <math> E</math> and <math> E_1</math> dnote the areas of triangles <math> ABC</math> and <math> AKL</math> respectively, show that
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\[ \frac {E}{E_1} \geq 2.
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<cmath> \frac {E}{E_1} \geq 2.</cmath>
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\]
 
  
 
==Solution==
 
==Solution==
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== See Also == {{IMO box|year=1988|num-b=4|num-a=6}}
 
== See Also == {{IMO box|year=1988|num-b=4|num-a=6}}
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 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 10:38, 30 January 2021

Problem

In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that \[\frac {E}{E_1} \geq 2.\]

Solution

Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then: \[\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC\] Proof of the lemma: Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:

$\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$

$=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$

If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$, $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$. Plug them in we get: \[\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC\]

Back to the problem Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get: \[\frac{AD}{AE} \cdot AB + \frac{AB}{AK} \cdot AD = AB+BD+AD\] and \[\frac{AD}{AE} \cdot AC + \frac{AC}{AL} \cdot AD = AC+CD+AD\] Cancel out the term $\frac{AD}{AE}$, we get: \[\frac{AB+BD+AD-\frac{AB}{AK} \cdot AD }{AC+CD+AD- \frac{AC}{AL} \cdot AD }=\frac{AB}{AC}\] \[AB \cdot CD + AB \cdot AD - \frac{AB \cdot AC \cdot AD}{AL}=AC \cdot BD+ AC \cdot AD -\frac{AB \cdot AC \cdot AD}{AK}\] \[AB+AB \cdot \frac{CD}{AD}-\frac{AB \cdot AC}{AL}=AC+ AC \cdot \frac{BD}{AD} - \frac{AB \cdot AC}{AK}\] \[AB+AC - \frac{AB \cdot AC}{AL}=AB+AC - \frac{AB \cdot AC}{AK}\] \[\frac{AB \cdot AC}{AK} = \frac{AB \cdot AC}{AL}\] So we conclude $AK=AL$.

Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$, thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is: \[\frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2\]

This solution was posted and copyrighted by shobber. The original thread for this problem can be found here: [1]

See Also

1988 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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