Difference between revisions of "1964 IMO Problems/Problem 6"
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== Problem == | == Problem == | ||
− | In tetrahedron <math>ABCD</math>, vertex <math>D</math> is connected with <math>D_0</math>, the | + | In tetrahedron <math>ABCD</math>, vertex <math>D</math> is connected with <math>D_0</math>, the centroid of <math>\triangle ABC</math>. Lines parallel to <math>DD_0</math> are drawn through <math>A,B</math> and <math>C</math>. These lines intersect the planes <math>BCD, CAD</math> and <math>ABD</math> in points <math>A_1, B_1,</math> and <math>C_1</math>, respectively. Prove that the volume of <math>ABCD</math> is one third the volume of <math>A_1B_1C_1D_0</math>. Is the result true if point <math>D_o</math> is selected anywhere within <math>\triangle ABC</math>? |
== Solution == | == Solution == | ||
+ | |||
+ | [[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>B_{2}</math> be the point where line <math>BD_{0}</math> intersects line <math>AC</math> | ||
+ | |||
+ | Let <math>C_{2}</math> be the point where line <math>CD_{0}</math> intersects line <math>AB</math> | ||
+ | |||
+ | From centroid properties we have: | ||
+ | |||
+ | <math>|AA_{2}|=3|D_{0}A_{2}|</math> | ||
+ | |||
+ | <math>|BB_{2}|=3|D_{0}B_{2}|</math> | ||
+ | |||
+ | <math>|CC_{2}|=3|D_{0}C_{2}|</math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>\frac{|AA_{2}|}{|D_{0}A_{2}|}=3</math> | ||
+ | |||
+ | <math>\frac{|BB_{2}|}{|D_{0}B_{2}|}=3</math> | ||
+ | |||
+ | <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_{1}|=|DD_{0}| \frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|</math> | ||
+ | |||
+ | Since <math>\Delta D_{0}B_{2}B_{1}\sim \Delta BB_{2}B_{1}</math>, then <math>|BB_{1}|=|DD_{0}| \frac{|BB_{2}|}{|D_{0}B_{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> | ||
+ | |||
+ | 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>, 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_{\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> | ||
+ | |||
+ | <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>h_{\Delta A_{1}B_{1}C_{1}}=3h_{D}</math> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | then, <math>Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}</math> | ||
+ | |||
+ | thus, <math>Volume_{ABCD}=\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}</math> | ||
+ | |||
+ | this proves that the volume of <math>ABCD</math> is one third the volume of <math>A_1B_1C_1D_0</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | == See Also == | ||
+ | {{IMO box|year=1964|num-b=5|after=Last Question}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 21:26, 21 November 2023
Problem
In tetrahedron , vertex
is connected with
, the centroid of
. Lines parallel to
are drawn through
and
. These lines intersect the planes
and
in points
and
, respectively. Prove that the volume of
is one third the volume of
. Is the result true if point
is selected anywhere within
?
Solution
Let be the point where line
intersects line
Let be the point where line
intersects line
Let be the point where line
intersects line
From centroid properties we have:
Therefore,
Since , then
Since , then
Since , then
Since and
,
then , and
Let be the perpendicular distance from
to
Let be the perpendicular distance from
to
Since and
then,
thus,
this proves that the volume of is one third the volume of
The result is NOT true if point is selected anywhere within
as ratios of
,
, and
will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios
which means that
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 | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |