Difference between revisions of "1992 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
Consider the plane through <math>A,A',B,B'</math>. This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired. | Consider the plane through <math>A,A',B,B'</math>. This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired. | ||
<math>\mathbb{QED.}</math> | <math>\mathbb{QED.}</math> | ||
− | == | + | ==== Add-on==== |
+ | By another person ^v^ | ||
− | {{ | + | The person that came up with the solution did not prove that <math>\triangle APB</math> is isosceles nor the base angles are congruent. I will add on to the solution. |
− | + | ||
+ | There is a common tangent plane that pass through <math>P</math> for the <math>2</math> spheres that are tangent to each other. | ||
+ | |||
+ | <br/> | ||
+ | Since any cross section of sphere is a circle. It implies that <math>A</math>, <math>A'</math>, <math>B</math>, <math>B'</math> be on the same circle (<math>\omega_1</math>), <math>A</math>, <math>B</math>, <math>P</math> be on the same circle (<math>\omega_2</math>), and <math>A'</math>, <math>B'</math>, <math>P</math> be on the same circle (<math>\omega_3</math>). | ||
+ | |||
+ | <math>m\angle APB= m\angle A'PB'</math> because they are vertical angles. By power of point, <math>(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}</math> | ||
+ | |||
+ | By the SAS triangle simlarity theory, <math>\triangle APB \sim\triangle B'PA'</math>. That implies that <math>\angle ABP\cong\angle B'PA'</math>. | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | Let's call the interception of the common tangent plane and the plane containing <math>A</math>, <math>A'</math>, <math>B</math>, <math>B'</math>, <math>P</math>, line <math>l</math>. | ||
+ | |||
+ | <math>l</math> must be the common tangent of <math>\omega_2</math> and <math>\omega_3</math>. | ||
+ | |||
+ | The acute angles form by <math>l</math> and <math>\overline{AA'}</math> are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord <math>\overline{AP}</math> and <math>\overline{A'P}</math> are equal. | ||
+ | |||
+ | Similarly the central angle of chord <math>\overline{BP}</math> and <math>\overline{B'P}</math> are equal. | ||
+ | |||
+ | The length of any chord with central angle <math>2\theta</math> and radius <math>r</math> is <math>2r\sin\left({\theta}\right)</math>, which can easily been seen if we drop the perpendicular from the center to the chord. | ||
+ | |||
+ | Thus, <math>\frac{AP}{A'P}=\frac{BP}{B'P}</math>. | ||
+ | |||
+ | By the SAS triangle simlarity theory, <math>\triangle APB \sim\triangle A'PB'</math>. That implies that <math>\angle ABP\cong\angle B'PA'</math>. | ||
+ | |||
+ | <br/> | ||
+ | That implies that <math>\angle ABP\cong\angle A'PB'\cong\angle B'PA'</math>. Thus, <math>\triangle A'PB'</math> is an isosceles triangle and since <math>\triangle APB \sim\triangle A'PB'</math>, then<math>\triangle APB</math> must be an isosceles triangle too. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | |||
+ | Call the large sphere <math> O_1</math>, the one containing <math>A</math> <math> O_2</math>, and the one containing <math> A'</math> <math>O_3</math>. The centers are <math> c_1</math>, <math> c_2,</math> and <math> c_3</math>. | ||
+ | |||
+ | |||
+ | Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle (<math> w_1</math>)completely contained in <math> O_1</math> and <math> O_2</math> | ||
+ | |||
+ | |||
+ | Similarly, A', B', and C' must lie on a circle (<math>w_2</math>) completely contained in <math> O_1</math> and <math> O_3</math>. | ||
+ | |||
+ | |||
+ | So, we know that 3 lines connecting a point on <math> w_1</math> and P hit a point on <math> w_2</math>. This implies that <math> w_1</math> projects through P to <math> w_2</math>, which in turn means that <math> w_1</math> is in a plane parallel to that of <math> w_2</math>. Then, since <math> w_1</math> and <math> w_2</math> lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center). | ||
+ | |||
+ | |||
+ | So, all line from a point on <math> w_1</math> to P are of the same length, as are all lines from a point on <math> w_2</math> to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal. | ||
+ | == See Also == | ||
− | [[Category:Olympiad | + | {{USAMO box|year=1992|num-b=3|num-a=5}} |
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 17:37, 25 June 2021
Problem
Chords ,
, and
of a sphere meet at an interior point
but are not contained in the same plane. The sphere through
,
,
, and
is tangent to the sphere through
,
,
, and
. Prove that
.
Solution
Solution 1
Consider the plane through . This plane, of course, also contains
. We can easily find the
is isosceles because the base angles are equal. Thus,
. Similarly,
. Thus,
. By symmetry,
and
, and hence
as desired.
Add-on
By another person ^v^
The person that came up with the solution did not prove that is isosceles nor the base angles are congruent. I will add on to the solution.
There is a common tangent plane that pass through for the
spheres that are tangent to each other.
Since any cross section of sphere is a circle. It implies that ,
,
,
be on the same circle (
),
,
,
be on the same circle (
), and
,
,
be on the same circle (
).
because they are vertical angles. By power of point,
By the SAS triangle simlarity theory, . That implies that
.
Let's call the interception of the common tangent plane and the plane containing ,
,
,
,
, line
.
must be the common tangent of
and
.
The acute angles form by and
are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord
and
are equal.
Similarly the central angle of chord and
are equal.
The length of any chord with central angle and radius
is
, which can easily been seen if we drop the perpendicular from the center to the chord.
Thus, .
By the SAS triangle simlarity theory, . That implies that
.
That implies that . Thus,
is an isosceles triangle and since
, then
must be an isosceles triangle too.
Solution 2
Call the large sphere , the one containing
, and the one containing
. The centers are
,
and
.
Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle ()completely contained in
and
Similarly, A', B', and C' must lie on a circle () completely contained in
and
.
So, we know that 3 lines connecting a point on and P hit a point on
. This implies that
projects through P to
, which in turn means that
is in a plane parallel to that of
. Then, since
and
lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center).
So, all line from a point on to P are of the same length, as are all lines from a point on
to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal.
See Also
1992 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.