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Difference between revisions of "1964 IMO Problems/Problem 6"

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== Solution ==
 
== Solution ==
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[[File:IMO_1964_P6_01.png]]
  
 
Let <math>A_{2}</math> be the point where line <math>AD_{0}</math> intersects line <math>BC</math>
 
Let <math>A_{2}</math> be the point where line <math>AD_{0}</math> intersects line <math>BC</math>
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Since <math>\Delta D_{0}C_{2}C_{1}\sim \Delta CC_{2}C_{1}</math>, then <math>|CC_{1}|=|DD_{0}| \frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|</math>
 
Since <math>\Delta D_{0}C_{2}C_{1}\sim \Delta CC_{2}C_{1}</math>, then <math>|CC_{1}|=|DD_{0}| \frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|</math>
  
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Since <math>|AA_{2}|=|BB_{2}|=|CC_{2}|</math> and <math>AA_{1} \parallel BB_{1} \parallel CC_{1} \parallel DD_{0}</math>, then <math>\Delta A_{1}B_{1}C_{1}\parallel \Delta ABC</math>
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Since <math>|AA_{2}|=|BB_{2}|=|CC_{2}|</math> and <math>AA_{1} \parallel BB_{1} \parallel CC_{1} \parallel DD_{0}</math>,
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then <math>\Delta A_{1}B_{1}C_{1}\parallel \Delta ABC</math>, and <math>Area_{\Delta A_{1}B_{1}C_{1}}=Area_{\Delta ABC}</math>
  
 
Let <math>h_{D}</math> be the perpendicular distance from <math>D</math> to <math>\Delta ABC</math>
 
Let <math>h_{D}</math> be the perpendicular distance from <math>D</math> to <math>\Delta ABC</math>
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Let <math>h_{\Delta A_{1}B_{1}C_{1}}</math> be the perpendicular distance from <math>\Delta A_{1}B_{1}C_{1}</math> to <math>\Delta ABC</math>
 
Let <math>h_{\Delta A_{1}B_{1}C_{1}}</math> be the perpendicular distance from <math>\Delta A_{1}B_{1}C_{1}</math> to <math>\Delta ABC</math>
  
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<math>frac{h_{\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\frac{|AA_{2}|}{|D_{0}A_{2}|}=\frac{|BB_{2}|}{|D_{0}B_{2}|}=\frac{|CC_{2}|}{|D_{0}C_{2}|}=3</math>
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<math>\frac{h_{\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\frac{|AA_{2}|}{|D_{0}A_{2}|}=\frac{|BB_{2}|}{|D_{0}B_{2}|}=\frac{|CC_{2}|}{|D_{0}C_{2}|}=3</math>
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<math>h_{\Delta A_{1}B_{1}C_{1}}=3h_{D}</math>
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<math>\frac{1}{3}h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}=3\frac{1}{3}h_{D}Area_{\Delta ABC}</math>
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<math>\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}=3\frac{h_{D}Area_{\Delta ABC}}{3}</math>
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Since <math>Volume_{ABCD}=\frac{h_{D}Area_{\Delta ABC}}{3}</math> and <math>Volume_{A_{1}B_{1}C_{1}D_{0}}=\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}</math>
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then, <math>Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}</math>
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thus, <math>Volume_{ABCD}=\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}</math>
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this proves that the volume of <math>ABCD</math> is one third the volume of <math>A_1B_1C_1D_0</math>
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The result is NOT true if point <math>D_o</math> is selected anywhere within <math>\triangle ABC</math> as ratios of <math>\frac{|AA_{2}|}{|D_{0}A_{2}|}</math>, <math>\frac{|BB_{2}|}{|D_{0}B_{2}|}</math>, and <math>\frac{|CC_{2}|}{|D_{0}C_{2}|}</math> will have values other than 3 as the point is no longer a centroid.  Also, it will make the ratios <math>\frac{|AA_{2}|}{|D_{0}A_{2}|} \ne \frac{|BB_{2}|}{|D_{0}B_{2}|}\ne \frac{|CC_{2}|}{|D_{0}C_{2}|}</math> which means that <math>\Delta A_{1}B_{1}C_{1} \nparallel \Delta ABC</math> and the volume relationship will no longer hold true.
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~Tomas Diaz. orders@tomasdiaz.com
  
  

Latest revision as of 20:26, 21 November 2023

Problem

In tetrahedron $ABCD$, vertex $D$ is connected with $D_0$, the centroid of $\triangle ABC$. Lines parallel to $DD_0$ are drawn through $A,B$ and $C$. These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_1, B_1,$ and $C_1$, respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$. Is the result true if point $D_o$ is selected anywhere within $\triangle ABC$?

Solution

IMO 1964 P6 01.png

Let $A_{2}$ be the point where line $AD_{0}$ intersects line $BC$

Let $B_{2}$ be the point where line $BD_{0}$ intersects line $AC$

Let $C_{2}$ be the point where line $CD_{0}$ intersects line $AB$

From centroid properties we have:

$|AA_{2}|=3|D_{0}A_{2}|$

$|BB_{2}|=3|D_{0}B_{2}|$

$|CC_{2}|=3|D_{0}C_{2}|$

Therefore,

$\frac{|AA_{2}|}{|D_{0}A_{2}|}=3$

$\frac{|BB_{2}|}{|D_{0}B_{2}|}=3$

$\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$

Since $\Delta D_{0}A_{2}A_{1}\sim \Delta AA_{2}A_{1}$, then $|AA_{1}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|$

Since $\Delta D_{0}B_{2}B_{1}\sim \Delta BB_{2}B_{1}$, then $|BB_{1}|=|DD_{0}| \frac{|BB_{2}|}{|D_{0}B_{2}|}=3|DD_{0}|$

Since $\Delta D_{0}C_{2}C_{1}\sim \Delta CC_{2}C_{1}$, then $|CC_{1}|=|DD_{0}| \frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|$

Since $|AA_{2}|=|BB_{2}|=|CC_{2}|$ and $AA_{1} \parallel BB_{1} \parallel CC_{1} \parallel DD_{0}$,

then $\Delta A_{1}B_{1}C_{1}\parallel \Delta ABC$, and $Area_{\Delta A_{1}B_{1}C_{1}}=Area_{\Delta ABC}$

Let $h_{D}$ be the perpendicular distance from $D$ to $\Delta ABC$

Let $h_{\Delta A_{1}B_{1}C_{1}}$ be the perpendicular distance from $\Delta A_{1}B_{1}C_{1}$ to $\Delta ABC$

$\frac{h_{\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\frac{|AA_{2}|}{|D_{0}A_{2}|}=\frac{|BB_{2}|}{|D_{0}B_{2}|}=\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$

$h_{\Delta A_{1}B_{1}C_{1}}=3h_{D}$

$\frac{1}{3}h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}=3\frac{1}{3}h_{D}Area_{\Delta ABC}$

$\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}=3\frac{h_{D}Area_{\Delta ABC}}{3}$

Since $Volume_{ABCD}=\frac{h_{D}Area_{\Delta ABC}}{3}$ and $Volume_{A_{1}B_{1}C_{1}D_{0}}=\frac{h_{\Delta A_{1}B_{1}C_{1}}Area_{\Delta A_{1}B_{1}C_{1}}}{3}$

then, $Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}$

thus, $Volume_{ABCD}=\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}$

this proves that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$

The result is NOT true if point $D_o$ is selected anywhere within $\triangle ABC$ as ratios of $\frac{|AA_{2}|}{|D_{0}A_{2}|}$, $\frac{|BB_{2}|}{|D_{0}B_{2}|}$, and $\frac{|CC_{2}|}{|D_{0}C_{2}|}$ will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios $\frac{|AA_{2}|}{|D_{0}A_{2}|} \ne \frac{|BB_{2}|}{|D_{0}B_{2}|}\ne \frac{|CC_{2}|}{|D_{0}C_{2}|}$ which means that $\Delta A_{1}B_{1}C_{1} \nparallel \Delta ABC$ and the volume relationship will no longer hold true.


~Tomas Diaz. orders@tomasdiaz.com


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1964 IMO (Problems) • Resources
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Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
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All IMO Problems and Solutions
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