Difference between revisions of "Power of a point theorem"
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==Theorem:== | ==Theorem:== | ||
Line 28: | Line 26: | ||
=====Classic Configuration===== | =====Classic Configuration===== | ||
− | Given lines <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(" | + | label("B",(2,3)); |
− | label(" | + | 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(" | + | label("D",(-3.350, -0.6)); |
dot((-2.429,1.761)); | dot((-2.429,1.761)); | ||
− | label(" | + | 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 59: | Line 57: | ||
draw((2.5,-1.658)---(-8,3)); | draw((2.5,-1.658)---(-8,3)); | ||
dot((-2.907,0.741)); | dot((-2.907,0.741)); | ||
− | label("D",(- | + | label("D",(-3.357,0.421)); |
</asy> | </asy> | ||
===Case 3 (On the Border/Useless Case):=== | ===Case 3 (On the Border/Useless Case):=== | ||
− | If two chords, <math> AB </math> and <math> AC </math>, 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 <math> 0 </math> so no matter what, the constant product is <math> 0 </math>. | + | If two chords, <math> AB </math> and <math> AC </math>, have <math> A </math> 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 <math> 0 </math> so no matter what, the constant product is <math> 0 </math>. |
+ | |||
+ | <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== | ==Proof== | ||
+ | |||
+ | ===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)=== | ||
+ | |||
+ | 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)=== | ||
+ | |||
+ | Length of a point is zero so no proof needed :) | ||
==Problems== | ==Problems== | ||
====Introductory (AMC 10, 12)==== | ====Introductory (AMC 10, 12)==== | ||
+ | |||
+ | Let <math>\overline{AB}</math> be a diameter in a circle of radius <math>5\sqrt2.</math> Let <math>\overline{CD}</math> be a chord in the circle that intersects <math>\overline{AB}</math> at a point <math>E</math> such that <math>BE=2\sqrt5</math> and <math>\angle AEC = 45^{\circ}.</math> What is <math>CE^2+DE^2?</math> | ||
+ | |||
+ | Source: [[2020 AMC 12B Problems/Problem 12]] | ||
====Intermediate (AIME)==== | ====Intermediate (AIME)==== | ||
+ | |||
+ | Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>. | ||
+ | |||
+ | Source: [[2024 AIME I Problems/Problem 10]] | ||
====Olympiad (USAJMO, USAMO, IMO)==== | ====Olympiad (USAJMO, USAMO, IMO)==== | ||
+ | |||
+ | Given circles <math>\omega_1</math> and <math>\omega_2</math> intersecting at points <math>X</math> and <math>Y</math>, let <math>\ell_1</math> be a line through the center of <math>\omega_1</math> intersecting <math>\omega_2</math> at points <math>P</math> and <math>Q</math> and let <math>\ell_2</math> be a line through the center of <math>\omega_2</math> intersecting <math>\omega_1</math> at points <math>R</math> and <math>S</math>. Prove that if <math>P, Q, R</math> and <math>S</math> lie on a circle then the center of this circle lies on line <math>XY</math>. | ||
+ | |||
+ | Source: [[2009 USAMO Problems/Problem 1]] | ||
+ | |||
+ | Let <math>P</math> be a point interior to triangle <math>ABC</math> (with <math>CA \neq CB</math>). The lines <math>AP</math>, <math>BP</math> and <math>CP</math> meet again its circumcircle <math>\Gamma</math> at <math>K</math>, <math>L</math>, respectively <math>M</math>. The tangent line at <math>C</math> to <math>\Gamma</math> meets the line <math>AB</math> at <math>S</math>. Show that from <math>SC = SP</math> follows <math>MK = ML</math>. | ||
+ | |||
+ | Source: [[2010 IMO Problems/Problem 4]] | ||
+ | |||
+ | ==Builders== | ||
+ | |||
+ | =====Creator:===== | ||
+ | =====Proof Writer: ===== | ||
+ | =====Other Editors (feel free to put username if you contributed):===== |
Latest revision as of 16:44, 28 June 2024
Contents
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 and
intersect at a point
within a circle, then
Case 2 (Outside the Circle):
Classic Configuration
Given lines and
originate from two unique points on the circumference of a circle (
and
), intersect each other at point
, outside the circle, and re-intersect the circle at points
and
respectively, then
Tangent Line
Given Lines and
with
tangent to the related circle at
,
lies outside the circle, and Line
intersects the circle between
and
at
,
Case 3 (On the Border/Useless Case):
If two chords, and
, have
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
so no matter what, the constant product is
.
Proof
Case 1 (Inside the Circle)
Join and
.
In
(Angles subtended by the same segment are equal)
(Vertically opposite angles)
(Corresponding sides of similar triangles are in the same ratio)
Case 2 (Outside the Circle)
Join and
(Why?)
Now, In
(shown above)
(common angle)
(Corresponding sides of similar triangles are in the same ratio)
Case 3 (On the Circle Border)
Length of a point is zero so no proof needed :)
Problems
Introductory (AMC 10, 12)
Let be a diameter in a circle of radius
Let
be a chord in the circle that intersects
at a point
such that
and
What is
Source: 2020 AMC 12B Problems/Problem 12
Intermediate (AIME)
Let be a triangle inscribed in circle
. Let the tangents to
at
and
intersect at point
, and let
intersect
at
. If
,
, and
,
can be written as the form
, where
and
are relatively prime integers. Find
.
Source: 2024 AIME I Problems/Problem 10
Olympiad (USAJMO, USAMO, IMO)
Given circles and
intersecting at points
and
, let
be a line through the center of
intersecting
at points
and
and let
be a line through the center of
intersecting
at points
and
. Prove that if
and
lie on a circle then the center of this circle lies on line
.
Source: 2009 USAMO Problems/Problem 1
Let be a point interior to triangle
(with
). The lines
,
and
meet again its circumcircle
at
,
, respectively
. The tangent line at
to
meets the line
at
. Show that from
follows
.
Source: 2010 IMO Problems/Problem 4