Difference between revisions of "1983 USAMO Problems/Problem 4"
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In this solution, we employ several lemmas. Two we shall take for granted: given any point <math>A</math> and a line <math>\ell</math> not passing through <math>A</math>, we can construct a line <math>\ell'</math> through <math>A</math> parallel to <math>\ell</math>; and given any point <math>A</math> on a line <math>\ell</math>, we can construct a line <math>\ell'</math> through <math>A</math> perpendicular to <math>\ell</math>. | In this solution, we employ several lemmas. Two we shall take for granted: given any point <math>A</math> and a line <math>\ell</math> not passing through <math>A</math>, we can construct a line <math>\ell'</math> through <math>A</math> parallel to <math>\ell</math>; and given any point <math>A</math> on a line <math>\ell</math>, we can construct a line <math>\ell'</math> through <math>A</math> perpendicular to <math>\ell</math>. | ||
− | + | '''Lemma 1:''' If we have two segments <math>S</math> and <math>T</math> on the plane with non-zero length, we may construct a circle at either endpoint of <math>S</math> whose radius is <math>|T|</math>. | |
− | ''' | ||
'''Proof:''' We can construct arbitrarily many copies of <math>T</math> by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of <math>T</math> like this until we create a circle of radius <math>|T|</math> and center <math>P_1</math> that intersects segment <math>S</math>. We can then take this intersection point <math>P_2</math> and draw a line <math>\ell</math> through it perpendicular to <math>S</math>, and draw a circle with center <math>P_2</math> passing through <math>P_1</math>, and consider its intersection <math>P_3</math> with <math>\ell</math>. Note that <math>P_2P_3\perp S</math> and <math>|P_2P_3|=|T|</math>. Take an endpoint <math>P_4</math> of <math>S</math>: then draw a line through <math>P_3</math> parallel to <math>S</math>, and a line through <math>P_4</math> parallel to <math>P_2P_3</math>. Let these two lines intersect at <math>P_5</math>. Then <math>P_2P_3P_5P_4</math> is a rectangle, so <math>|P_4P_5|=|T|</math>. Our desired circle is then a circle centered at <math>P_4</math> through <math>P_5</math>. | '''Proof:''' We can construct arbitrarily many copies of <math>T</math> by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of <math>T</math> like this until we create a circle of radius <math>|T|</math> and center <math>P_1</math> that intersects segment <math>S</math>. We can then take this intersection point <math>P_2</math> and draw a line <math>\ell</math> through it perpendicular to <math>S</math>, and draw a circle with center <math>P_2</math> passing through <math>P_1</math>, and consider its intersection <math>P_3</math> with <math>\ell</math>. Note that <math>P_2P_3\perp S</math> and <math>|P_2P_3|=|T|</math>. Take an endpoint <math>P_4</math> of <math>S</math>: then draw a line through <math>P_3</math> parallel to <math>S</math>, and a line through <math>P_4</math> parallel to <math>P_2P_3</math>. Let these two lines intersect at <math>P_5</math>. Then <math>P_2P_3P_5P_4</math> is a rectangle, so <math>|P_4P_5|=|T|</math>. Our desired circle is then a circle centered at <math>P_4</math> through <math>P_5</math>. | ||
− | + | '''Lemma 2:''' Given three collinear points <math>A</math>, <math>B</math>, <math>C</math> in this order, if <math>|AB|=a</math> and <math>|BC|=b</math> with <math>a>b</math>, then we can construct a segment of length <math>\sqrt{a^2-b^2}</math>. | |
− | ''' | ||
'''Proof:''' From Lemma 1, we can construct a circle through <math>C</math> with radius <math>a</math>, and then construct a perpendicular through <math>B</math> to <math>AC</math>: these two objects intersect at <math>D</math> and <math>E</math>. Both <math>BD</math> and <math>DE</math> have length <math>\sqrt{a^2-b^2}</math>, from the Pythagorean Theorem. | '''Proof:''' From Lemma 1, we can construct a circle through <math>C</math> with radius <math>a</math>, and then construct a perpendicular through <math>B</math> to <math>AC</math>: these two objects intersect at <math>D</math> and <math>E</math>. Both <math>BD</math> and <math>DE</math> have length <math>\sqrt{a^2-b^2}</math>, from the Pythagorean Theorem. | ||
− | + | '''Proof of the original statement:''' Note that we can construct a triangle <math>B'C'D'</math> congruent to triangle <math>BCD</math> by applying Lemma 1 to segments <math>S_4</math>, <math>S_5</math>, and <math>S_6</math>. Similarly, we can construct <math>A_B</math> and <math>A_C</math> outside triangle <math>B'C'D'</math> such that <math>A_BC'D'\cong ACD</math> and <math>A_CB'D'\cong ABD</math>. | |
− | Note that we can construct a triangle <math>B'C'D'</math> congruent to triangle <math>BCD</math> by applying Lemma 1 to segments <math>S_4</math>, <math>S_5</math>, and <math>S_6</math>. Similarly, we can construct <math>A_B</math> and <math>A_C</math> outside triangle <math>B'C'D'</math> such that <math>A_BC'D'\cong ACD</math> and <math>A_CB'D'\cong ABD</math>. | ||
Let <math>A'</math> be a point outside of the plane containing <math>S_1</math> through <math>S_6</math> such that <math>A'B'C'D'\cong ABCD</math>. Then the altitudes of triangles <math>A_BC'D'</math> and <math>A'C'D'</math> to segment <math>C'D'</math> are congruent, as are the altitudes of triangles <math>A_CB'D'</math> and <math>A'B'D'</math> to segment <math>B'D'</math>. However, if we project the altitudes of <math>A'C'D'</math> and <math>A'B'D'</math> from <math>A'</math> onto the plane, their intersection is the base of the altitude of tetrahedron <math>A'B'C'D'</math> from <math>A'</math>. In addition, these altitude projections are collinear with the altitudes of triangles <math>A_BC'D'</math> and <math>A_CB'D'</math>. Therefore, the altitudes of <math>A_BC'D'</math> and <math>A_CB'D'</math> from <math>A_B</math> and <math>A_C</math> intersect at the base <math>X'</math> of the altitude of <math>A'B'C'D'</math> from <math>A'</math>. In summary, we can construct <math>X'</math> by constructing the perpendiculars from <math>A_B</math> and <math>A_C</math> to <math>C'D'</math> and <math>B'D'</math> respectively, and taking their intersection. | Let <math>A'</math> be a point outside of the plane containing <math>S_1</math> through <math>S_6</math> such that <math>A'B'C'D'\cong ABCD</math>. Then the altitudes of triangles <math>A_BC'D'</math> and <math>A'C'D'</math> to segment <math>C'D'</math> are congruent, as are the altitudes of triangles <math>A_CB'D'</math> and <math>A'B'D'</math> to segment <math>B'D'</math>. However, if we project the altitudes of <math>A'C'D'</math> and <math>A'B'D'</math> from <math>A'</math> onto the plane, their intersection is the base of the altitude of tetrahedron <math>A'B'C'D'</math> from <math>A'</math>. In addition, these altitude projections are collinear with the altitudes of triangles <math>A_BC'D'</math> and <math>A_CB'D'</math>. Therefore, the altitudes of <math>A_BC'D'</math> and <math>A_CB'D'</math> from <math>A_B</math> and <math>A_C</math> intersect at the base <math>X'</math> of the altitude of <math>A'B'C'D'</math> from <math>A'</math>. In summary, we can construct <math>X'</math> by constructing the perpendiculars from <math>A_B</math> and <math>A_C</math> to <math>C'D'</math> and <math>B'D'</math> respectively, and taking their intersection. | ||
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== See Also == | == See Also == | ||
− | {{USAMO box|year= | + | {{USAMO box|year=1983|num-b=3|num-a=5}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
+ | [[Category:Geometric Construction Problems]] |
Latest revision as of 09:58, 19 July 2016
Problem
Six segments and
are given in a plane. These are congruent to the edges
and
, respectively, of a tetrahedron
. Show how to construct a segment congruent to the altitude of the tetrahedron from vertex
with straight-edge and compasses.
Solution
Throughout this solution, we denote the length of a segment by
.
In this solution, we employ several lemmas. Two we shall take for granted: given any point and a line
not passing through
, we can construct a line
through
parallel to
; and given any point
on a line
, we can construct a line
through
perpendicular to
.
Lemma 1: If we have two segments and
on the plane with non-zero length, we may construct a circle at either endpoint of
whose radius is
.
Proof: We can construct arbitrarily many copies of by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of
like this until we create a circle of radius
and center
that intersects segment
. We can then take this intersection point
and draw a line
through it perpendicular to
, and draw a circle with center
passing through
, and consider its intersection
with
. Note that
and
. Take an endpoint
of
: then draw a line through
parallel to
, and a line through
parallel to
. Let these two lines intersect at
. Then
is a rectangle, so
. Our desired circle is then a circle centered at
through
.
Lemma 2: Given three collinear points ,
,
in this order, if
and
with
, then we can construct a segment of length
.
Proof: From Lemma 1, we can construct a circle through with radius
, and then construct a perpendicular through
to
: these two objects intersect at
and
. Both
and
have length
, from the Pythagorean Theorem.
Proof of the original statement: Note that we can construct a triangle congruent to triangle
by applying Lemma 1 to segments
,
, and
. Similarly, we can construct
and
outside triangle
such that
and
.
Let be a point outside of the plane containing
through
such that
. Then the altitudes of triangles
and
to segment
are congruent, as are the altitudes of triangles
and
to segment
. However, if we project the altitudes of
and
from
onto the plane, their intersection is the base of the altitude of tetrahedron
from
. In addition, these altitude projections are collinear with the altitudes of triangles
and
. Therefore, the altitudes of
and
from
and
intersect at the base
of the altitude of
from
. In summary, we can construct
by constructing the perpendiculars from
and
to
and
respectively, and taking their intersection.
Let be the intersection of
with
. Then the altitude length we seek to construct is, from the Pythagorean Theorem,
. We can directly apply Lemma 2 to segment
to obtain this segment. This shows how to construct a segment of length
.
See Also
1983 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.