Difference between revisions of "1992 USAMO Problems/Problem 4"

Latest revision as of 17:37, 25 June 2021

Problem

Chords $AA'$ , $BB'$ , and $CC'$ of a sphere meet at an interior point $P$ but are not contained in the same plane. The sphere through $A$ , $B$ , $C$ , and $P$ is tangent to the sphere through $A'$ , $B'$ , $C'$ , and $P$ . Prove that $AA'=BB'=CC'$ .

Solution

Solution 1

Consider the plane through $A,A',B,B'$ . This plane, of course, also contains $P$ . We can easily find the $\triangle APB$ is isosceles because the base angles are equal. Thus, $AP=BP$ . Similarly, $A'P=B'P$ . Thus, $AA'=BB'$ . By symmetry, $BB'=CC'$ and $CC'=AA'$ , and hence $AA'=BB'=CC'$ as desired.

$\mathbb{QED.}$

Add-on

By another person ^v^

The person that came up with the solution did not prove that $\triangle APB$ is isosceles nor the base angles are congruent. I will add on to the solution.

There is a common tangent plane that pass through $P$ for the $2$ spheres that are tangent to each other.

Since any cross section of sphere is a circle. It implies that $A$ , $A'$ , $B$ , $B'$ be on the same circle ( $\omega_1$ ), $A$ , $B$ , $P$ be on the same circle ( $\omega_2$ ), and $A'$ , $B'$ , $P$ be on the same circle ( $\omega_3$ ).

$m\angle APB= m\angle A'PB'$ because they are vertical angles. By power of point, $(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}$

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle B'PA'$ . That implies that $\angle ABP\cong\angle B'PA'$ .

Let's call the interception of the common tangent plane and the plane containing $A$ , $A'$ , $B$ , $B'$ , $P$ , line $l$ .

$l$ must be the common tangent of $\omega_2$ and $\omega_3$ .

The acute angles form by $l$ and $\overline{AA'}$ are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\overline{AP}$ and $\overline{A'P}$ are equal.

Similarly the central angle of chord $\overline{BP}$ and $\overline{B'P}$ are equal.

The length of any chord with central angle $2\theta$ and radius $r$ is $2r\sin\left({\theta}\right)$ , which can easily been seen if we drop the perpendicular from the center to the chord.

Thus, $\frac{AP}{A'P}=\frac{BP}{B'P}$ .

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle A'PB'$ . That implies that $\angle ABP\cong\angle B'PA'$ .

That implies that $\angle ABP\cong\angle A'PB'\cong\angle B'PA'$ . Thus, $\triangle A'PB'$ is an isosceles triangle and since $\triangle APB \sim\triangle A'PB'$ , then $\triangle APB$ must be an isosceles triangle too.

Solution 2

Call the large sphere $O_1$ , the one containing $A$ $O_2$ , and the one containing $A'$ $O_3$ . The centers are $c_1$ , $c_2,$ and $c_3$ .

Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle ( $w_1$ )completely contained in $O_1$ and $O_2$

Similarly, A', B', and C' must lie on a circle ( $w_2$ ) completely contained in $O_1$ and $O_3$ .

So, we know that 3 lines connecting a point on $w_1$ and P hit a point on $w_2$ . This implies that $w_1$ projects through P to $w_2$ , which in turn means that $w_1$ is in a plane parallel to that of $w_2$ . Then, since $w_1$ and $w_2$ 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 $w_1$ to P are of the same length, as are all lines from a point on $w_2$ to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal.

@@ Line 3: / Line 3: @@
 ==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.
 <math>\mathbb{QED.}</math>
-== Resources ==
+==== Add-on====
+By another person ^v^
-{{USAMO box|year=1992|num-b=3|num-a=5}}
+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.
-* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks]
+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 Algebra Problems]]
+{{USAMO box|year=1992|num-b=3|num-a=5}}
+{{MAA Notice}}
+[[Category:Olympiad Geometry Problems]]
+[[Category:3D Geometry Problems]]

1992 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5
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