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Difference between revisions of "1969 Canadian MO Problems/Problem 4"

 
 
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== Problem ==
 
== Problem ==
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Let <math>\displaystyle ABC</math> be an equilateral triangle, and <math>\displaystyle P</math> be an arbitrary point within the triangle. Perpendiculars <math>\displaystyle PD,PE,PF</math> are drawn to the three sides of the triangle. Show that, no matter where <math>\displaystyle P</math> is chosen, <math>\displaystyle \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>.
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Let <math>ABC</math> be an equilateral triangle, and <math>P</math> be an arbitrary point within the triangle. Perpendiculars <math>PD,PE,PF</math> are drawn to the three sides of the triangle. Show that, no matter where <math>P</math> is chosen, <math>\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>.
  
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== Solution ==
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== Solution 1 ==
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Let a side of the triangle be <math>\displaystyle s</math> and let <math>\displaystyle [ABC]</math> denote the area of <math>\displaystyle ABC.</math> Note that because <math>\displaystyle 2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],</math> <math>\displaystyle\frac{\sqrt3}{2}s^2=s(PD+PE+PF).</math> Dividing both sides by <math>\displaystyle s</math>, the sum of the perpendiculars from <math>\displaystyle P</math> equals <math>PD+PE+PF=\frac{\sqrt3}{2}s.</math> (It is independant of point <math>\displaystyle P</math>) Because the sum of the sides is <math>\displaystyle 3s</math>, the ratio is always <math>\displaystyle\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.</math>
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Let a side of the triangle be <math>s</math> and let <math>[ABC]</math> denote the area of <math>ABC.</math> Note that because <math>2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],</math> <math>\frac{\sqrt3}{2}s^2=s(PD+PE+PF).</math> Dividing both sides by <math>s</math>, the sum of the perpendiculars from <math>P</math> equals <math>PD+PE+PF=\frac{\sqrt3}{2}s.</math> (It is independant of point <math>P</math>) Because the sum of the sides is <math>3s</math>, the ratio is always <math>\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.</math>
  
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== Solution 2 ==
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* [[1969 Canadian MO Problems/Problem 3|Previous Problem]]
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* [[1969 Canadian MO Problems/Problem 5|Next Problem]]
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Let a be the side of the triangle. By Viviani's theorem (which has a similar proof to solution 1) PD + PE + PF = h where h is the height of the equilateral triangle. Now it is not hard to see that the ratio h : 3 * a = 1 / 2 root 3  from 30 - 60 - 90 triangles or trigonometry.  ~AK2006
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* [[1969 Canadian MO Problems|Back to Exam]]
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{{Old CanadaMO box|num-b=3|num-a=5|year=1969}}

Latest revision as of 23:23, 8 October 2020

Problem

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, $\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}$.

Solution 1

Let a side of the triangle be $s$ and let $[ABC]$ denote the area of $ABC.$ Note that because $2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],$ $\frac{\sqrt3}{2}s^2=s(PD+PE+PF).$ Dividing both sides by $s$, the sum of the perpendiculars from $P$ equals $PD+PE+PF=\frac{\sqrt3}{2}s.$ (It is independant of point $P$) Because the sum of the sides is $3s$, the ratio is always $\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.$

Solution 2

Let a be the side of the triangle. By Viviani's theorem (which has a similar proof to solution 1) PD + PE + PF = h where h is the height of the equilateral triangle. Now it is not hard to see that the ratio h : 3 * a = 1 / 2 root 3 from 30 - 60 - 90 triangles or trigonometry. ~AK2006

1969 Canadian MO (Problems)
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 5


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