Difference between revisions of "Power of a Point Theorem"
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− | The '''Power of a Point Theorem''' expresses the relation between the lengths involved with the intersection of two | + | The '''Power of a Point Theorem''' expresses the relation between the lengths involved with the intersection of two [[line]]s between each other and their [[intersection]]s 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. | ||
− | # The two lines are [[secant line|secants]] of the circle and intersect inside the circle (figure on the left). | + | # The two lines are [[secant line|secants]] 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). | + | # 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> | # 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> | ||
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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> | <cmath> | ||
PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i | PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i | ||
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[[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). | + | 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 == | ||
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The theorem generalizes to higher dimensions, as follows. | 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 | + | 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> | <cmath> | ||
− | PA_1 \cdot PB_1 = PA_2 \cdot PB_2 | + | PA_1\cdot PB_1=PA_2\cdot PB_2 |
</cmath> | </cmath> | ||
− | ''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. | + | ''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 == | ||
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=== 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]] | [[Image:popprob1.PNG|center]] | ||
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==== 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]] | [[Image:popprob2.PNG|center]] | ||
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==== 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]] | [[Image:popprob3.PNG|center]] | ||
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==== 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. 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]] | ||
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=== 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. Find the perimeter of <math>\triangle PQR</math>. |
==== Other Intermediate Example Problems ==== | ==== Other Intermediate Example Problems ==== |
Revision as of 11:16, 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.
Contents
Theorem
There are three possibilities as displayed in the figures below.
- The two lines are secants 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
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. 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. Find the perimeter of
.