Difference between revisions of "Power of a Point Theorem"

Latest revision as of 09:36, 16 February 2025

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.

Statement

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

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 $PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i$

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 any 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 (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

Introductory

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

Solution

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

Solution

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

Solution

(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

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 $\triangle PQR$ . (Source)

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. (Source)

$DEB$ is a chord of a circle such that $DE=3$ and $EB=5 .$ Let $O$ be the center of the circle. Join $OE$ and extend $OE$ to cut the circle at $C.$ Given $EC=1,$ find the radius of the circle. (Source)

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ (Source)

Olympiad

Template:Incomplete

@@ Line 1: / Line 1: @@
 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 ==
+== Statement ==
-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|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>
+# 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.)
@@ Line 24: / Line 24: @@
 == 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.
+One important result of this theorem is that both tangents from any 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>
+<cmath>PA_1\cdot PB_1=PA_2\cdot PB_2</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.
+''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (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 ==
-The problems are divided into three categories: introductory, intermediate, and olympiad.
 === Introductory ===
-==== Problem 1 ====
-Find the value of <math>x</math> in the following diagram:
-[[Image:popprob1.PNG|center]]
+* Find the value of <math>x</math> in the following diagram: [[Image:popprob1.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
-[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
+* Find the value of <math>x</math> in the following diagram: [[Image:popprob2.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
-==== Problem 2 ====
+* ([[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>   .
-Find the value of <math>x</math> in the following diagram:
+[[Image:popprob3.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
-[[Image:popprob2.PNG|center]]
+* ([[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_2|Solution]]
+=== Intermediate ===
-==== Problem 3 ====
+* 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|Source]])
-([[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]]
+* 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. ([[2020 AMC 12B Problems/Problem 10|Source]])
-[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
+* <math>DEB</math> is a chord of a circle such that <math>DE=3</math> and <math>EB=5 .</math> Let <math>O</math> be the center of the circle. Join <math>OE</math> and extend <math>OE</math> to cut the circle at <math>C.</math> Given <math>EC=1,</math> find the radius of the circle. ([[1971_Canadian_MO_Problems/Problem_1|Source]])
+[[Image:CanadianMO_1971-1.jpg]]
-==== Problem 4 ====
+* Triangle <math>ABC</math> has <math>BC=20.</math> The incircle of the triangle evenly trisects the median <math>AD.</math> If the area of the triangle is <math>m \sqrt{n}</math> where <math>m</math> and <math>n</math> are integers and <math>n</math> is not divisible by the square of a prime, find <math>m+n.</math> ([[2005 AIME I Problems/Problem 15|Source]])
-([[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]]
-=== Intermediate ===
-==== 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>.
-==== Problem 2 ====
+=== Olympiad ===
-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 ====
+* {{incomplete}}
-* [[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]]
 * [[Planar figures]]
+* [{{SERVER}}/community/c6h2513977 Kagebaka's Handout]
 [[Category:Geometry]]
 [[Category:Theorems]]
+{{stub}}

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Difference between revisions of "Power of a Point Theorem"

Latest revision as of 09:36, 16 February 2025

Contents

Statement

Hint for Proof

Alternate Formulation

Additional Notes

Problems

Introductory

Intermediate

Olympiad

See Also