Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Power of a Point Theorem"

m
(Categories; minor edits)
Line 1: Line 1:
 
The '''Power of a Point Theorem''' expresses the relation between the lengths involved with the intersection of two lines between each other and their intersections with a [[circle]].
 
The '''Power of a Point Theorem''' expresses the relation between the lengths involved with the intersection of two lines between each other and their intersections with a [[circle]].
−
 
  
 
== Theorem ==
 
== Theorem ==
−
 
 
There are three possibilities as displayed in the figures below.
 
There are three possibilities as displayed in the figures below.
  
Line 10: Line 8:
 
# Both lines are [[secant line|secants]] of the circle and intersect outside of it (figure on the right).  In this case, we have <math> CB\cdot CA = CD\cdot CE. </math>
 
# Both lines are [[secant line|secants]] of the circle and intersect outside of it (figure on the right).  In this case, we have <math> CB\cdot CA = CD\cdot CE. </math>
  
−
<center>[[Image:Pop.PNG]]</center>
+
[[Image:Pop.PNG|center]]
−
 
 
  
 
=== Alternate Formulation ===
 
=== Alternate Formulation ===
Line 17: Line 14:
  
 
Consider a circle O and a point P in the plane where P is not on the circle.  Now draw a line through P that intersects the circle in two places.  The power of a point theorem says that the product of the length from P to the first point of intersection and the length from P to the second point of intersection is constant for any choice of a line through P that intersects the circle.  This constant is called the power of point P.  For example, in the figure below  
 
Consider a circle O and a point P in the plane where P is not on the circle.  Now draw a line through P that intersects the circle in two places.  The power of a point theorem says that the product of the length from P to the first point of intersection and the length from P to the second point of intersection is constant for any choice of a line through P that intersects the circle.  This constant is called the power of point P.  For example, in the figure below  
 +
<cmath>
 +
PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i
 +
</cmath>
 +
[[Image:Popalt.PNG|center]]
  
−
<center><math> PX^2 = PA_1\cdot PB_1 = PA_2\cdot PB_2 = \cdots = PA_i\cdot PB_i </math>.</center>
+
Notice how this definition still works if <math>A_k</math> and <math>B_k</math> coincide (as is the case with X).  Consider also when P is inside the circle.  The definition still holds in this case.
−
 
 
−
<center>[[Image:Popalt.PNG]]</center>
 
−
 
 
−
Notice how this definition still works if <math> A_k </math> and <math> B_k </math> coincide (as is the case with X).  Consider also when P is inside the circle.  The definition still holds in this case.
 
−
 
 
  
 
== Additional Notes ==
 
== Additional Notes ==
Line 30: Line 26:
 
The theorem generalizes to higher dimensions, as follows.
 
The theorem generalizes to higher dimensions, as follows.
  
−
Let <math>\displaystyle P </math> be a point, and let <math>\displaystyle S </math> be an <math>\displaystyle n </math>-sphere. Let two arbitrary lines passing through <math>\displaystyle P </math> intersect <math>\displaystyle S </math> at <math>\displaystyle A_1 , B_1 ; A_2 , B_2 </math>, respectively. Then
+
Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P </math> intersect <math>S </math> at <math>A_1 , B_1 ; A_2 , B_2 </math>, respectively. Then
 +
<cmath>
 +
PA_1 \cdot PB_1 = PA_2 \cdot PB_2
 +
</cmath>
  
−
<center>
+
''Proof.''  We have already proven the theorem for a 1-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] <math>p </math> containing both of the lines passing through <math>P </math>.  The intersection of <math>P </math> and <math>S </math> must be a circle.  If we consider the lines and <math>P </math> with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
−
<math>\displaystyle PA_1 \cdot PB_1 = PA_2 \cdot PB_2 </math>
 
−
</center>
 
−
 
 
−
''Proof.''  We have already proven the theorem for a 1-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] <math>\displaystyle p </math> containing both of the lines passing through <math>\displaystyle P </math>.  The intersection of <math>\displaystyle P </math> and <math>\displaystyle S </math> must be a circle.  If we consider the lines and <math>\displaystyle P </math> with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
 
  
 
== Problems ==
 
== Problems ==
Line 45: Line 40:
 
Find the value of <math> x </math> in the following diagram:
 
Find the value of <math> x </math> in the following diagram:
  
−
<center>[[Image:popprob1.PNG]]</center>
+
[[Image:popprob1.PNG|center]]
  
 
[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
 
[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
Line 52: Line 47:
 
Find the value of <math> x </math> in the following diagram.
 
Find the value of <math> x </math> in the following diagram.
  
−
<center>[[Image:popprob2.PNG]]</center>
+
[[Image:popprob2.PNG|center]]
  
 
[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
 
[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
Line 59: Line 54:
 
([[ARML]]) In a circle, chords <math> AB </math> and <math> CD </math> intersect at <math> R </math>. If <math>AR:BR = 1:4 </math> and <math> CR:DR = 4:9 </math>, find the ratio <math> AB:CD. </math>
 
([[ARML]]) In a circle, chords <math> AB </math> and <math> CD </math> intersect at <math> R </math>. If <math>AR:BR = 1:4 </math> and <math> CR:DR = 4:9 </math>, find the ratio <math> AB:CD. </math>
  
−
<center>[[Image:popprob3.PNG]]</center>
+
[[Image:popprob3.PNG|center]]
  
 
[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
 
[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
Line 74: Line 69:
 
==== Other Intermediate Example Problems ====
 
==== Other Intermediate Example Problems ====
 
* [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]]
 
* [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]]
−
 
−
 
−
{{problems}}
 
  
 
==See also==
 
==See also==
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Planar figures]]
 
* [[Planar figures]]
 +
 +
[[Category:Geometry]]
 +
[[Category:Theorems]]

Revision as of 11:12, 27 April 2008

The Power of a Point Theorem expresses the relation between the lengths involved with the intersection of two lines between each other and their intersections with a circle.

Theorem

There are three possibilities as displayed in the figures below.

  1. The two lines are secants of the circle and intersect inside the circle (figure on the left). In this case, we have $AE\cdot CE = BE\cdot DE$.
  2. One of the lines is tangent to the circle while the other is a secant (middle figure). In this case, we have $AB^2 = BC\cdot BD$.
  3. Both lines are secants of the circle and intersect outside of it (figure on the right). In this case, we have $CB\cdot CA = CD\cdot CE.$
Pop.PNG

Alternate Formulation

This alternate formulation is much more compact, convenient, and general.

Consider a circle O and a point P in the plane where P is not on the circle. Now draw a line through P that intersects the circle in two places. The power of a point theorem says that the product of the length from P to the first point of intersection and the length from P to the second point of intersection is constant for any choice of a line through P that intersects the circle. This constant is called the power of point P. For example, in the figure below \[PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i\]

Popalt.PNG

Notice how this definition still works if $A_k$ and $B_k$ coincide (as is the case with X). Consider also when P is inside the circle. The definition still holds in this case.

Additional Notes

One important result of this theorem is that both tangents from a point $P$ outside of a circle to that circle are equal in length.

The theorem generalizes to higher dimensions, as follows.

Let $P$ be a point, and let $S$ be an $n$-sphere. Let two arbitrary lines passing through $P$ intersect $S$ at $A_1 , B_1 ; A_2 , B_2$, respectively. Then \[PA_1 \cdot PB_1 = PA_2 \cdot PB_2\]

Proof. We have already proven the theorem for a 1-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions. Consider the plane $p$ containing both of the lines passing through $P$. The intersection of $P$ and $S$ must be a circle. If we consider the lines and $P$ with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.

Problems

The problems are divided into three categories: introductory, intermediate, and olympiad.

Introductory

Problem 1

Find the value of $x$ in the following diagram:

Popprob1.PNG

Solution

Problem 2

Find the value of $x$ in the following diagram.

Popprob2.PNG

Solution

Problem 3

(ARML) In a circle, chords $AB$ and $CD$ intersect at $R$. If $AR:BR = 1:4$ and $CR:DR = 4:9$, find the ratio $AB:CD.$

Popprob3.PNG

Solution

Problem 4

(ARML) Chords $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle. Given that $BE = 16, DE = 4,$ and $AD = 5$, find $CE$.

Solution

Intermediate

Problem 1

Two tangents from an external point $P$ are drawn to a circle and intersect it at $A$ and $B$. A third tangent meets the circle at $T$, and the tangents $\overrightarrow{PA}$ and $\overrightarrow{PB}$ at points $Q$ and $R$, respectively. Find the perimeter of $\triangle PQR$.

Other Intermediate Example Problems

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Power_of_a_Point_Theorem&oldid=25615"