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Difference between revisions of "Power of a point theorem"

(→‎Proof)
(→‎Case 3 (On the Circle Border))
 
(6 intermediate revisions by 2 users not shown)
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=====Classic Configuration=====
 
=====Classic Configuration=====
  
−
Given lines <math> AB </math> and <math> CB </math> originate from two unique points on the [[circumference]] of a circle (<math> A </math> and <math> C </math>), intersect each other at point <math> B </math>, outside the circle, and re-intersect the circle at points <math> F </math> and <math> G </math> respectively, then <math> BF\cdot BA=BG\cdot BC </math>
+
Given lines <math> BP </math> and <math> CP </math> originate from two unique points on the [[circumference]] of a circle (<math> B </math> and <math> C </math>), intersect each other at point <math> P </math>, outside the circle, and re-intersect the circle at points <math> A </math> and <math> D </math> respectively, then <math> PA\cdot PB=PD\cdot PC </math>
  
 
<asy> draw(circle((0,0),3));  
 
<asy> draw(circle((0,0),3));  
 
dot((1.5,2.598));
 
dot((1.5,2.598));
−
label("A",(2,3));
+
label("B",(2,3));
−
label("B",(-6,1.6));
+
label("P",(-6,1.6));
 
dot((-6,1));
 
dot((-6,1));
 
label("C",(2.55,-2.5));
 
label("C",(2.55,-2.5));
 
dot((2.12,-2.123));
 
dot((2.12,-2.123));
 
dot((-2.996,-0.155));
 
dot((-2.996,-0.155));
−
label("G",(-3.350, -0.6));
+
label("D",(-3.350, -0.6));
 
dot((-2.429,1.761));
 
dot((-2.429,1.761));
−
label("F",(-2.729,2.061));
+
label("A",(-2.729,2.061));
 
draw((1.5,2.598)---(-6,1));
 
draw((1.5,2.598)---(-6,1));
 
draw((2.12,-2.123)---(-6,1));
 
draw((2.12,-2.123)---(-6,1));
Line 77: Line 77:
 
==Proof==
 
==Proof==
  
−
Please write a proof of you have one and feel free to put name under proof writers at bottom of page :)!
 
 
===Case 1 (Inside the Circle)===
 
===Case 1 (Inside the Circle)===
 +
 +
Join <math>AD</math> and <math>BC</math>.
 +
 +
In <math>\triangle ADP \; \text{and} \; \triangle CBP</math>
 +
 +
<math>\angle ADC = \angle CBA \hspace{1cm}</math>  (Angles subtended by the same segment are equal)
 +
 +
<math>\angle DPA = \angle BPC \hspace{1cm}</math>  (Vertically opposite angles)
 +
 +
<math>\therefore \; \triangle ADP \sim \triangle CBP</math>
 +
 +
<math>\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
 +
 +
<math>\implies AP \cdot BP = DP \cdot CP</math>
 +
 +
<math>\blacksquare</math>
  
 
===Case 2 (Outside the Circle)===
 
===Case 2 (Outside the Circle)===
 +
 +
Join <math>AD</math> and <math>BC</math>
 +
 +
<math>\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}</math> (Why?)
 +
 +
<math>\implies \angle PCB = \angle DCB = \angle PAD</math>
 +
 +
Now, In <math>\triangle PAD \; \text{and} \; \triangle PCB</math>
 +
 +
<math>\angle PAD = \angle PCB \hspace{1cm}</math> (shown above)
 +
 +
<math>\angle APD = \angle CPB \hspace{1cm}</math> (common angle)
 +
 +
<math>\therefore \; \triangle PAD \sim \triangle PCB</math>
 +
 +
<math>\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
 +
 +
<math>\implies PA \cdot PB = PD \cdot PC</math>
 +
 +
<math>\blacksquare</math>
  
 
===Case 3 (On the Circle Border)===
 
===Case 3 (On the Circle Border)===
 +
 +
Length of a point is zero so no proof needed :)
  
 
==Problems==
 
==Problems==

Latest revision as of 15:44, 28 June 2024

Theorem:

There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application.

Case 1 (Inside the Circle):

If two chords $AB$ and $CD$ intersect at a point $P$ within a circle, then $AP\cdot BP=CP\cdot DP$

[asy] draw(circle((0,0),3)); dot((-2.82,1)); label("A",(-3.05,1.25)); dot((1,2.828)); label("B",(1.25,3.05)); draw((-2.82,1)---(1,2.828)); dot((2.3,-1.926)); label("C",(2.55,-2.346)); dot((-2.12,2.123)); label("D",(-2.37,2.507)); draw((2.3,-1.926)---(-2.12,2.123)); dot((-1.556,1.602)); label("P",(-1.656,1.202)); [/asy]

Case 2 (Outside the Circle):

Classic Configuration

Given lines $BP$ and $CP$ originate from two unique points on the circumference of a circle ($B$ and $C$), intersect each other at point $P$, outside the circle, and re-intersect the circle at points $A$ and $D$ respectively, then $PA\cdot PB=PD\cdot PC$

[asy] draw(circle((0,0),3)); dot((1.5,2.598)); label("B",(2,3)); label("P",(-6,1.6)); dot((-6,1)); label("C",(2.55,-2.5)); dot((2.12,-2.123)); dot((-2.996,-0.155)); label("D",(-3.350, -0.6)); dot((-2.429,1.761)); label("A",(-2.729,2.061)); draw((1.5,2.598)---(-6,1)); draw((2.12,-2.123)---(-6,1)); [/asy]

Tangent Line

Given Lines $AB$ and $AC$ with $AC$ tangent to the related circle at $C$, $A$ lies outside the circle, and Line $AB$ intersects the circle between $A$ and $B$ at $D$, $AD\cdot AB=AC^{2}$

[asy] draw(circle((0,0),3)); dot((0,3)); label("C",(0,3.5)); dot((-8,3)); label("A",(-8,3.5)); dot((2.5,-1.658)); label("B",(2.8,-1.958)); draw((0,3)---(-8,3)); draw((2.5,-1.658)---(-8,3)); dot((-2.907,0.741)); label("D",(-3.357,0.421)); [/asy]

Case 3 (On the Border/Useless Case):

If two chords, $AB$ and $AC$, have $A$ on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is $0$ so no matter what, the constant product is $0$.

[asy] draw(circle((0,0),3)); dot((1,2.828)); label("A",(1.4,3.028)); dot((-2.5,-1.658)); label("B",(-2.8,-1.958)); dot((2.04,-2.2)); label("C",(2.34,-2.5)); draw((1,2.828)---(-2.5,-1.658)); draw((1,2.828)---(2.04,-2.2)); [/asy]

Proof

Case 1 (Inside the Circle)

Join $AD$ and $BC$.

In $\triangle ADP \; \text{and} \; \triangle CBP$

$\angle ADC = \angle CBA \hspace{1cm}$ (Angles subtended by the same segment are equal)

$\angle DPA = \angle BPC \hspace{1cm}$ (Vertically opposite angles)

$\therefore \; \triangle ADP \sim \triangle CBP$

$\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies AP \cdot BP = DP \cdot CP$

$\blacksquare$

Case 2 (Outside the Circle)

Join $AD$ and $BC$

$\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}$ (Why?)

$\implies \angle PCB = \angle DCB = \angle PAD$

Now, In $\triangle PAD \; \text{and} \; \triangle PCB$

$\angle PAD = \angle PCB \hspace{1cm}$ (shown above)

$\angle APD = \angle CPB \hspace{1cm}$ (common angle)

$\therefore \; \triangle PAD \sim \triangle PCB$

$\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies PA \cdot PB = PD \cdot PC$

$\blacksquare$

Case 3 (On the Circle Border)

Length of a point is zero so no proof needed :)

Problems

Introductory (AMC 10, 12)

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$

Source: 2020 AMC 12B Problems/Problem 12

Intermediate (AIME)

Let $ABC$ be a triangle inscribed in circle $\omega$. Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$, and let $\overline{AD}$ intersect $\omega$ at $P$. If $AB=5$, $BC=9$, and $AC=10$, $AP$ can be written as the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers. Find $m + n$.

Source: 2024 AIME I Problems/Problem 10

Olympiad (USAJMO, USAMO, IMO)

Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.

Source: 2009 USAMO Problems/Problem 1

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Source: 2010 IMO Problems/Problem 4

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