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Difference between revisions of "Power of a Point Theorem"

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== Statement ==
 
== Statement ==
 +
 
There are three possibilities as displayed in the figures below:
 
There are three possibilities as displayed in the figures below:
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# 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>.
+
# The two lines are [[chord]]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>.
−
# 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>.
+
# 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]]
 
[[Image:Pop.PNG|center]]
  
−
=== Hint for Proof===
+
=== Alternate Formulation ===
−
Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another line as well.)
 
  
−
=== 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 <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  
 
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</cmath>
−
PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i
 
−
</cmath>
 
 
[[Image:Popalt.PNG|center]]
 
[[Image:Popalt.PNG|center]]
 +
 +
=== Hint for Proof===
 +
 +
Draw extra lines to create similar triangles (Draw <math>AD</math> on all three figures. Draw another line as well.)
  
 
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.
 
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 ==
+
== Notes ==
 +
 
 
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.
 
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.
  
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<cmath>PA_1\cdot PB_1=PA_2\cdot PB_2</cmath>
 
<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 (a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] p 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 ==
 
== Problems ==
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=== Olympiad ===
 
=== Olympiad ===
  
−
* {{incomplete}}
+
* {{problem}}
  
 
== See Also ==
 
== See Also ==
 +
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Planar figures]]
 
* [[Planar figures]]

Latest revision as of 16:04, 21 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:

  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

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

Hint for Proof

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

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.

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:
    Popprob1.PNG
Solution
  • Find the value of $x$ in the following diagram:
    Popprob2.PNG
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 $AB:CD$ .
Popprob3.PNG
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)

CanadianMO 1971-1.jpg

  • 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

  • This problem has not been edited in. Help us out by adding it.

See Also

This article is a stub. Help us out by expanding it.

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