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

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<math>|CC_{2}|=3|D_{0}C_{2}|</math>
 
<math>|CC_{2}|=3|D_{0}C_{2}|</math>
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Therefore,
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<math>\frac{|AA_{2}|}{|D_{0}A_{2}|}=3</math>
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<math>\frac{|BB_{2}|}{|D_{0}B_{2}|}=3</math>
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<math>\frac{|CC_{2}|}{|D_{0}C_{2}|}=3</math>
  
 
Since <math>\Delta D_{0}A_{2}A_{1}\sim \Delta AA_{2}A_{1}</math>, then <math>|AA_{2}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}</math>
 
Since <math>\Delta D_{0}A_{2}A_{1}\sim \Delta AA_{2}A_{1}</math>, then <math>|AA_{2}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}</math>
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{{solution}}
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{{alternate solutions}}
  
 
== See Also ==  
 
== See Also ==  

Revision as of 23:48, 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 $|AA_{2}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}$


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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