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

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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>
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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> and <math>Area_{\Delta A_{1}B_{1}C_{1}}=Area_{\Delta ABC}</math>
  
 
<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>
 
<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>

Revision as of 23:00, 16 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

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 $\Delta A_{1}B_{1}C_{1}\parallel \Delta ABC$

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

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


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
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
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