Difference between revisions of "Power of a Point Theorem"
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− | + | 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. | |
− | The '''Power of a Point Theorem''' | ||
== Theorem == | == Theorem == | ||
+ | There are three possibilities as displayed in the figures below. | ||
− | + | # 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 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 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> | ||
− | + | [[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. | + | 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> | ||
+ | PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i | ||
+ | </cmath> | ||
+ | [[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 <math>X</math>). Consider also when <math>P</math> 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 <math> P </math> outside of a circle to that circle are equal in length. | ||
− | + | The theorem generalizes to higher dimensions, as follows. | |
− | + | 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> | ||
+ | ''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. | ||
== Problems == | == Problems == | ||
Line 32: | 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: |
− | + | [[Image:popprob1.PNG|center]] | |
− | [[Power of a Point Theorem | + | [[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: |
− | + | [[Image:popprob2.PNG|center]] | |
− | [[Power of a Point Theorem | + | [[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 | + | ([[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> . |
− | + | [[Image:popprob3.PNG|center]] | |
− | [[Power of a Point | + | [[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. | + | ([[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 | + | [[Power of a Point Theorem/Introductory_Problem_4|Solution]] |
=== 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. | + | 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 ==== | |
+ | * [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]] | ||
+ | * [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]] | ||
==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 14:30, 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.
Contents
Theorem
There are three possibilities as displayed in the figures below.
- The two lines are chords of the circle and intersect inside the circle (figure on the left). In this case, we have
.
- One of the lines is tangent to the circle while the other is a secant (middle figure). In this case, we have
.
- Both lines are secants of the circle and intersect outside of it (figure on the right). In this case, we have
Hint for Proof
Draw extra lines to create similar triangles! (Hint: Draw on all three figures. Draw another line as well.)
Alternate Formulation
This alternate formulation is much more compact, convenient, and general.
Consider a circle and a point
in the plane where
is not on the circle. Now draw a line through
that intersects the circle in two places. The power of a point theorem says that the product of the length from
to the first point of intersection and the length from
to the second point of intersection is constant for any choice of a line through
that intersects the circle. This constant is called the power of point
. For example, in the figure below
Notice how this definition still works if and
coincide (as is the case with
). Consider also when
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 outside of a circle to that circle are equal in length.
The theorem generalizes to higher dimensions, as follows.
Let be a point, and let
be an
-sphere. Let two arbitrary lines passing through
intersect
at
, respectively. Then
Proof. We have already proven the theorem for a -sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions. Consider the plane
containing both of the lines passing through
. The intersection of
and
must be a circle. If we consider the lines and
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 in the following diagram:
Problem 2
Find the value of in the following diagram:
Problem 3
(ARML) In a circle, chords and
intersect at
. If
and
, find the ratio
.
Problem 4
(ARML) Chords and
of a given circle are perpendicular to each other and intersect at a right angle at point
. Given that
,
, and
, find
.
Intermediate
Problem 1
Two tangents from an external point are drawn to a circle and intersect it at
and
. A third tangent meets the circle at
, and the tangents
and
at points
and
, respectively (this means that T is on the minor arc
). If
, find the perimeter of
.
Problem 2
Square of side length
has a circle inscribed in it. Let
be the midpoint of
. Find the length of that portion of the segment
that lies outside of the circle.