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"

(→‎Additional Notes: added note about higher dimensions)
m (→‎Hint for Proof)
 
(39 intermediate revisions by 25 users not shown)
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''' is a relationship that holds between the lengths of the [[line segment]]s formed when two [[line]]s [[intersect]] a [[circle]] and each other.
−
 
 
  
 
== Theorem ==
 
== Theorem ==
−
 
 
There are three possibilities as displayed in the figures below.
 
There are three possibilities as displayed in the figures below.
  
−
# The two lines are [[secant]]s of the circle and intersect inside the circle (figure on the left). In this case, we have <math> AE\cdot CE = BE\cdot DE </math>.
+
# The two lines are [[secant line|chords]] of the circle and intersect inside the circle (figure on the left). In this case, we have <math> AE\cdot CE = BE\cdot DE </math>.
−
# One of the lines is [[tangent]] to the circle while the other is a [[secant]] (middle figure). In this case, we have <math> AB^2 = BC\cdot BD </math>.
+
# One of the lines is [[tangent line|tangent]] to the circle while the other is a [[secant line|secant]] (middle figure). In this case, we have <math> AB^2 = BC\cdot BD </math>.
−
# Both lines are [[secant]]s 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]]
 +
=== Hint for Proof===
 +
Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another line as well.)
  
 
=== Alternate Formulation ===
 
=== Alternate Formulation ===
 
This alternate formulation is much more compact, convenient, and general.
 
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  
+
Consider a circle <math>O</math> and a point <math>P</math> in the plane where <math>P</math> is not on the circle. Now draw a line through <math>P</math> that intersects the circle in two places. The power of a point theorem says that the product of the length from <math>P</math> to the first point of intersection and the length from <math>P</math> to the second point of intersection is constant for any choice of a line through <math>P</math> that intersects the circle. This constant is called the power of point <math>P</math>. For example, in the figure below  
−
 
+
<cmath>
−
<center><math> PX^2 = PA_1\cdot PB_1 = PA_2\cdot PB_2 = \cdots = PA_i\cdot PB_i </math>.</center>
+
PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i
−
 
+
</cmath>
−
<center>[[Image:Popalt.PNG]]</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.
 
  
 +
Notice how this definition still works if <math>A_k</math> and <math>B_k</math> coincide (as is the case with <math>X</math>). Consider also when <math>P</math> is inside the circle. The definition still holds in this case.
  
 
== Additional Notes ==
 
== Additional Notes ==
Line 30: Line 28:
 
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 <math>1</math>-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 43: Line 40:
 
=== Introductory ===
 
=== Introductory ===
 
==== Problem 1 ====
 
==== Problem 1 ====
−
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]]
  
 
==== Problem 2 ====   
 
==== Problem 2 ====   
−
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]]
  
 
==== Problem 3 ====  
 
==== Problem 3 ====  
−
([[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]]
  
 
==== Problem 4 ====
 
==== Problem 4 ====
−
([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle. Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>.
+
([[ARML]]) Chords <math>AB</math> and <math>CD</math> of a given circle are [[perpendicular]] to each other and intersect at a right angle at point <math>E</math>. Given that <math>BE=16</math>, <math>DE=4</math>, and <math>AD=5</math>, find <math>CE</math>.
  
 
[[Power of a Point Theorem/Introductory_Problem_4|Solution]]
 
[[Power of a Point Theorem/Introductory_Problem_4|Solution]]
Line 70: Line 67:
 
=== Intermediate ===
 
=== Intermediate ===
 
==== Problem 1 ====
 
==== Problem 1 ====
−
Two tangents from an external point <math> P </math> are drawn to a circle and intersect it at <math> A </math> and <math> B </math>.  A third tangent meets the circle at <math> T </math>, and the tangents <math> \overrightarrow{PA} </math> and <math> \overrightarrow{PB} </math> at points <math> Q </math> and <math> R </math>, respectively. Find the perimeter of <math> \triangle PQR </math>.
+
Two tangents from an external point <math>P</math> are drawn to a circle and intersect it at <math>A</math> and <math>B</math>.  A third tangent meets the circle at <math>T</math>, and the tangents <math>\overrightarrow{PA}</math> and <math>\overrightarrow{PB}</math> at points <math>Q</math> and <math>R</math>, respectively (this means that T is on the minor arc <math>AB</math>). If <math>AP = 20</math>, find the perimeter of <math>\triangle PQR</math>.
 +
 
 +
[[1961_AHSME_Problems/Problem_11|Solution]]
 +
 
 +
==== Problem 2 ====
 +
Square <math>ABCD</math> of side length <math>10</math> has a circle inscribed in it. Let <math>M</math> be the midpoint of <math>\overline{AB}</math>. Find the length of that portion of the segment <math>\overline{MC}</math> that lies outside of the circle.
  
 
==== 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]]
−
 
+
* [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]]
−
 
 
−
{{problems}}
 
  
 
==See also==
 
==See also==
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Planar figures]]
 
* [[Planar figures]]
 +
* [https://artofproblemsolving.com/community/c6h2513977 Kagebaka's Handout]
 +
[[Category:Geometry]]
 +
[[Category:Theorems]]

Latest revision as of 13:30, 30 June 2024

The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two lines intersect a circle and each other.

Theorem

There are three possibilities as displayed in the figures below.

  1. The two lines are chords 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

Hint for Proof

Draw extra lines to create similar triangles! (Hint: Draw $AD$ on all three figures. Draw another line as well.)

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 at point $E$. 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 (this means that T is on the minor arc $AB$). If $AP = 20$, find the perimeter of $\triangle PQR$.

Solution

Problem 2

Square $ABCD$ of side length $10$ has a circle inscribed in it. Let $M$ be the midpoint of $\overline{AB}$. Find the length of that portion of the segment $\overline{MC}$ that lies outside of the circle.

Other Intermediate Example Problems

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Power_of_a_Point_Theorem&oldid=221352"