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Difference between revisions of "1985 OIM Problems/Problem 6"
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== Solution ==
 
== Solution ==
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{{solution}}
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Drop an altitude from <math>A</math>, and let the intersection be <math>M</math>. Drop another perpendicular from <math>O</math> to <math>BC</math>, and let this intersection be <math>N</math>. Then notice that <math>\triangle AMD\sim\triangle OND</math>. Let <math>T</math> be the area of <math>\triangle ABC</math>; then <math>AM=\frac{2T}{a}</math>. By definition, <math>AO=r</math>. Let <math>AD=x</math>; finally, by definition, <math>ON</math> is the perpendicular bisector of <math>\overline{BC}</math>, so by the [[Pythagorean Theorem]],
 +
<cmath>ON^2=OB^2-BN^2</cmath>
 +
But <math>OB</math> is a radius and thus is equal to <math>r</math> (hence <math>OD=x-r</math> below), and <math>BN=\frac{1}{2}a</math> from the perpendicular bisector, so
 +
<cmath>ON=\sqrt{r^2-\frac{1}{4}a^2}</cmath>
 +
Then, by similarity,
 +
<cmath>\frac{AM}{AD}=\frac{ON}{OD}</cmath>
 +
<cmath>\iff\frac{\frac{2T}{a}}{x}=\frac{\sqrt{r^2-\frac{1}{4}a^2}}{x-r}</cmath>
 +
<cmath>\iff \frac{ax}{2T}=\frac{x-r}{\sqrt{r^2-\frac{1}{4}a^2}}</cmath>
 +
<cmath>\iff x\left(\frac{a}{2T}-\frac{1}{\sqrt{r^2-\frac{1}{4}a^2}}\right)=-\frac{r}{\sqrt{r^2-\frac{1}{4}a^2}}</cmath>
 +
<cmath>\iff x\left(1-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2T}\right)=r</cmath>
 +
<cmath>\iff x=\frac{2Tr}{2T-a\sqrt{r^2-\frac{1}{4}a^2}}</cmath>
 +
<cmath>\iff \frac{1}{x}=\frac{2T-a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}=\frac{1}{r}-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}</cmath>
 +
Then we must prove that
 +
<cmath>\frac{3}{r}-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}-\frac{b\sqrt{r^2-\frac{1}{4}b^2}}{2Tr}-\frac{c\sqrt{r^2-\frac{1}{4}c^2}}{2Tr}=\frac{2}{r}</cmath>
 +
which is equivalent to
 +
<cmath>\iff\sum_\text{cyc}\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}=\frac{1}{r}</cmath>
 +
or
 +
<cmath>\iff\sum_\text{cyc}a\sqrt{4r^2-a^2}=4T</cmath>
 +
But we note that <math>a=2r\sin\alpha</math> by the Extended [[Law of Sines]], so substituting:
 +
<cmath>\iff\sum_\text{cyc}a\sqrt{4r^2-4r^2\sin^2\alpha}=\sum_\text{cyc}a\sqrt{4r(1-\sin^2\alpha)}=\sum_\text{cyc}a\sqrt{4r^2\cos^2\alpha}=\sum_\text{cyc}2ar\cos\alpha=4T</cmath>
 +
with the last equality coming from noticing that <math>\cos\alpha</math> is nonnegative over <math>\alpha\in[0,\pi]</math>. Thus,
 +
<cmath>\iff\sum_\text{cyc}a\cos\alpha=\frac{2T}{r}</cmath>
 +
But from the [[Law of Cosines]], <math>\cos\alpha=\frac{b^2+c^2-a^2}{2bc}</math>, so
 +
<cmath>\iff\sum_\text{cyc}a\left(\frac{b^2+c^2-a^2}{2bc}\right)=\frac{2T}{r}</cmath>
 +
<cmath>\iff\sum_\text{cyc}a^2(b^2+c^2-a^2)=\frac{4abcT}{r}</cmath>
 +
Additionally, <math>T=\frac{abc}{4r}</math>, and expanding the left-hand side:
 +
<cmath>\iff2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4=16T^2=16s(s-a)(s-b)(s-c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)</cmath>
 +
using [[Heron's Formula]]. Expanding the right-hand side yields the conclusion.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe1.htm
 
https://www.oma.org.ar/enunciados/ibe1.htm

Latest revision as of 16:41, 19 April 2025

Problem

Given triangle $ABC$, we consider the points $D$, $E$, and $F$ of lines $BC$, $AC$, and $AB$ respectively. If lines $AD$, $BE$, and $CF$ all pass through the center $O$ of the circumference of triangle $ABC$, which radius is $r$, prove: \[\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CE}=\frac{2}{r}\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Drop an altitude from $A$, and let the intersection be $M$. Drop another perpendicular from $O$ to $BC$, and let this intersection be $N$. Then notice that $\triangle AMD\sim\triangle OND$. Let $T$ be the area of $\triangle ABC$; then $AM=\frac{2T}{a}$. By definition, $AO=r$. Let $AD=x$; finally, by definition, $ON$ is the perpendicular bisector of $\overline{BC}$, so by the Pythagorean Theorem, \[ON^2=OB^2-BN^2\] But $OB$ is a radius and thus is equal to $r$ (hence $OD=x-r$ below), and $BN=\frac{1}{2}a$ from the perpendicular bisector, so \[ON=\sqrt{r^2-\frac{1}{4}a^2}\] Then, by similarity, \[\frac{AM}{AD}=\frac{ON}{OD}\] \[\iff\frac{\frac{2T}{a}}{x}=\frac{\sqrt{r^2-\frac{1}{4}a^2}}{x-r}\] \[\iff \frac{ax}{2T}=\frac{x-r}{\sqrt{r^2-\frac{1}{4}a^2}}\] \[\iff x\left(\frac{a}{2T}-\frac{1}{\sqrt{r^2-\frac{1}{4}a^2}}\right)=-\frac{r}{\sqrt{r^2-\frac{1}{4}a^2}}\] \[\iff x\left(1-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2T}\right)=r\] \[\iff x=\frac{2Tr}{2T-a\sqrt{r^2-\frac{1}{4}a^2}}\] \[\iff \frac{1}{x}=\frac{2T-a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}=\frac{1}{r}-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}\] Then we must prove that \[\frac{3}{r}-\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}-\frac{b\sqrt{r^2-\frac{1}{4}b^2}}{2Tr}-\frac{c\sqrt{r^2-\frac{1}{4}c^2}}{2Tr}=\frac{2}{r}\] which is equivalent to \[\iff\sum_\text{cyc}\frac{a\sqrt{r^2-\frac{1}{4}a^2}}{2Tr}=\frac{1}{r}\] or \[\iff\sum_\text{cyc}a\sqrt{4r^2-a^2}=4T\] But we note that $a=2r\sin\alpha$ by the Extended Law of Sines, so substituting: \[\iff\sum_\text{cyc}a\sqrt{4r^2-4r^2\sin^2\alpha}=\sum_\text{cyc}a\sqrt{4r(1-\sin^2\alpha)}=\sum_\text{cyc}a\sqrt{4r^2\cos^2\alpha}=\sum_\text{cyc}2ar\cos\alpha=4T\] with the last equality coming from noticing that $\cos\alpha$ is nonnegative over $\alpha\in[0,\pi]$. Thus, \[\iff\sum_\text{cyc}a\cos\alpha=\frac{2T}{r}\] But from the Law of Cosines, $\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$, so \[\iff\sum_\text{cyc}a\left(\frac{b^2+c^2-a^2}{2bc}\right)=\frac{2T}{r}\] \[\iff\sum_\text{cyc}a^2(b^2+c^2-a^2)=\frac{4abcT}{r}\] Additionally, $T=\frac{abc}{4r}$, and expanding the left-hand side: \[\iff2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4=16T^2=16s(s-a)(s-b)(s-c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\] using Heron's Formula. Expanding the right-hand side yields the conclusion.

~ eevee9406

See also

https://www.oma.org.ar/enunciados/ibe1.htm

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