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| − | ==Theorem:==
| + | #REDIRECT [[Power of a Point Theorem]] |
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| − | 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.
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| − | ===Case 1 (Inside the Circle):===
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| − | If two chords <math> AB </math> and <math> CD </math> intersect at a point <math> P </math> within a circle, then <math> AP\cdot BP=CP\cdot DP </math>
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| − | <asy> draw(circle((0,0),3));
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| − | dot((-2.82,1));
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| − | label("A",(-3.05,1.25));
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| − | dot((1,2.828));
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| − | label("B",(1.25,3.05));
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| − | draw((-2.82,1)---(1,2.828));
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| − | dot((2.3,-1.926));
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| − | label("C",(2.55,-2.346));
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| − | dot((-2.12,2.123));
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| − | label("D",(-2.37,2.507));
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| − | draw((2.3,-1.926)---(-2.12,2.123));
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| − | dot((-1.556,1.602));
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| − | label("P",(-1.656,1.202));
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| − | </asy>
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| − | ===Case 2 (Outside the Circle):===
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| − | =====Classic Configuration=====
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| − | 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>
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| − | <asy> draw(circle((0,0),3));
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| − | dot((1.5,2.598));
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| − | label("B",(2,3));
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| − | label("P",(-6,1.6));
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| − | dot((-6,1));
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| − | label("C",(2.55,-2.5));
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| − | dot((2.12,-2.123));
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| − | dot((-2.996,-0.155));
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| − | label("D",(-3.350, -0.6));
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| − | dot((-2.429,1.761));
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| − | label("A",(-2.729,2.061));
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| − | draw((1.5,2.598)---(-6,1));
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| − | draw((2.12,-2.123)---(-6,1));
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| − | </asy>
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| − | =====Tangent Line=====
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| − | Given Lines <math> AB </math> and <math> AC </math> with <math> AC </math> [[tangent line|tangent]] to the related circle at <math> C </math>, <math> A </math> lies outside the circle, and Line <math> AB </math> intersects the circle between <math> A </math> and <math> B </math> at <math> D </math>, <math> AD\cdot AB=AC^{2} </math>
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| − | <asy> draw(circle((0,0),3));
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| − | dot((0,3));
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| − | label("C",(0,3.5));
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| − | dot((-8,3));
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| − | label("A",(-8,3.5));
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| − | dot((2.5,-1.658));
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| − | label("B",(2.8,-1.958));
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| − | draw((0,3)---(-8,3));
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| − | draw((2.5,-1.658)---(-8,3));
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| − | dot((-2.907,0.741));
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| − | label("D",(-3.357,0.421));
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| − | </asy>
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| − | ===Case 3 (On the Border/Useless Case):===
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| − | 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>.
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| − | <asy> draw(circle((0,0),3));
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| − | dot((1,2.828));
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| − | label("A",(1.4,3.028));
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| − | dot((-2.5,-1.658));
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| − | label("B",(-2.8,-1.958));
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| − | dot((2.04,-2.2));
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| − | label("C",(2.34,-2.5));
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| − | draw((1,2.828)---(-2.5,-1.658));
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| − | draw((1,2.828)---(2.04,-2.2));
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| − | </asy>
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| − | ==Proof==
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| − | ===Case 1 (Inside the Circle)===
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| − | Join <math>AD</math> and <math>BC</math>.
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| − | In <math>\triangle ADP \; \text{and} \; \triangle CBP</math>
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| − | <math>\angle ADC = \angle CBA \hspace{1cm}</math> (Angles subtended by the same segment are equal)
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| − | <math>\angle DPA = \angle BPC \hspace{1cm}</math> (Vertically opposite angles)
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| − | <math>\therefore \; \triangle ADP \sim \triangle CBP</math>
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| − | <math>\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
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| − | <math>\implies AP \cdot BP = DP \cdot CP</math>
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| − | <math>\blacksquare</math>
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| − | ===Case 2 (Outside the Circle)===
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| − | Join <math>AD</math> and <math>BC</math>
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| − | <math>\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}</math> (Why?)
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| − | <math>\implies \angle PCB = \angle DCB = \angle PAD</math>
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| − | Now, In <math>\triangle PAD \; \text{and} \; \triangle PCB</math>
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| − | <math>\angle PAD = \angle PCB \hspace{1cm}</math> (shown above)
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| − | <math>\angle APD = \angle CPB \hspace{1cm}</math> (common angle)
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| − | <math>\therefore \; \triangle PAD \sim \triangle PCB</math>
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| − | <math>\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
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| − | <math>\implies PA \cdot PB = PD \cdot PC</math>
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| − | <math>\blacksquare</math>
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| − | ===Case 3 (On the Circle Border)===
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| − | Length of a point is zero so no proof needed :)
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| − | ==Problems==
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| − | ====Introductory (AMC 10, 12)====
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| − | 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>
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| − | Source: [[2020 AMC 12B Problems/Problem 12]]
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| − | ====Intermediate (AIME)====
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| − | 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>.
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| − | Source: [[2024 AIME I Problems/Problem 10]]
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| − | ====Olympiad (USAJMO, USAMO, IMO)====
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| − | 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>.
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| − | Source: [[2009 USAMO Problems/Problem 1]]
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| − | 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>.
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| − | Source: [[2010 IMO Problems/Problem 4]]
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| − | ==Builders==
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| − | =====Creator:=====
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| − | =====Proof Writer: =====
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| − | =====Other Editors (feel free to put username if you contributed):=====
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