Difference between revisions of "Radical axis"
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==Definitions== | ==Definitions== | ||
− | The '''power''' of point <math>P</math> with respect to circle <math>\omega</math> (with radius <math>r</math> and center <math>O</math>), which shall thereafter be dubbed <math>pow(P, \omega)</math>, is defined to equal <math>OP^2 - r^2</math>. | + | The '''power''' of point <math>P</math> with respect to circle <math>\omega</math> (with radius <math>r</math> and center <math>O</math>), which shall thereafter be dubbed <math>pow(P, \omega)</math>, is defined to equal <math>|OP|^2 - r^2</math>. |
Note that the power of a point is negative if the point is inside the circle. | Note that the power of a point is negative if the point is inside the circle. | ||
− | The '''radical axis''' of two non-concentric circles <math>\omega_1, \omega_2</math> is defined as the locus of the points <math>P</math> such that the [[power of a point|power]] of <math>P</math> with respect to <math>\omega_1</math> and <math>\omega_2</math> are equal. In other words, if <math>O_i, r_i</math> are the center and radius of <math>\omega_i</math>, then a point <math>P</math> is on the radical axis if and only if <cmath>PO_1^2 - r_1^2 = PO_2^2 - r_2^2</cmath> | + | The '''radical axis''' of two non-concentric circles <math>\omega_1, \omega_2</math> is defined as the locus of the points <math>P</math> such that the [[power of a point|power]] of <math>P</math> with respect to <math>\omega_1</math> and <math>\omega_2</math> are equal. In other words, if <math>O_i, r_i</math> are the center and radius of <math>\omega_i</math>, then a point <math>P</math> is on the radical axis if and only if <cmath>|PO_1|^2 - r_1^2 = |PO_2|^2 - r_2^2</cmath> (i.e., the radical axis is the line that one gets when you subtract the equations of two circles). |
==Results== | ==Results== | ||
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d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to <math>l</math> through point A, the unique point on <math>l</math> such that <math>pow(A, \omega_1) = pow(A, \omega_2)</math>. | d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to <math>l</math> through point A, the unique point on <math>l</math> such that <math>pow(A, \omega_1) = pow(A, \omega_2)</math>. | ||
− | |||
− | |||
'''Theorem 3: (Radical Axis Concurrence Theorem)''' | '''Theorem 3: (Radical Axis Concurrence Theorem)''' | ||
The three pairwise radical axes of three circles concur at a point, called the '''radical center'''. | The three pairwise radical axes of three circles concur at a point, called the '''radical center'''. | ||
+ | |||
+ | '''Theorem 4: (Radical Centre of Intersecting Circles) (EGMO Theorem 2.9)''' | ||
+ | Let <math>\omega_1</math> and <math>\omega_2</math> be two circles with centers <math>O_1</math> and <math>O_2</math>. Select two points A and B on <math>\omega_1</math> and <math>\omega_2</math>. Then the following are equivalent | ||
+ | *<math>A,B,C,D</math> lie on a circle with center <math>O_3</math> not on line <math>O_1O_2</math>. | ||
+ | *Lines <math>AB</math> and <math>CD</math> intersect on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>. | ||
+ | |||
+ | '''Theorem 5: (EGMO Lemma 2.11)''' | ||
+ | Let ABC be a triangle and consider a point <math>P</math> in its interior. Suppose that <math>BC</math> is tangent to <math>(ABP)</math> and <math>(ACP)</math>, ray <math>AP</math> bisects <math>BC</math> | ||
==Proofs== | ==Proofs== | ||
− | Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.) | + | '''Theorem 1''' is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.) |
− | Theorem 2 shall be proved here. Assume the circles are <math>\omega_1</math> and <math>\omega_2</math> with centers <math>O_1</math> and <math>O_2</math> and radii <math>r_1</math> and <math>r_2</math>, respectively. (It may be a good idea for you to draw some circles here.) | + | '''Theorem 2''' shall be proved here. Assume the circles are <math>\omega_1</math> and <math>\omega_2</math> with centers <math>O_1</math> and <math>O_2</math> and radii <math>r_1</math> and <math>r_2</math>, respectively. (It may be a good idea for you to draw some circles here.) |
First, we tackle part (b). Suppose the circles intersect at points <math>X</math> and <math>Y</math> and point P lies on <math>XY</math>. Then by Theorem 1 the powers of P with respect to both circles are equal to <math>PX * PY</math>, and hence by transitive <math>pow(P, \omega_1) = pow(P, \omega_2)</math>. Thus, if point P lies on <math>XY</math>, then the powers of P with respect to both circles are equal. | First, we tackle part (b). Suppose the circles intersect at points <math>X</math> and <math>Y</math> and point P lies on <math>XY</math>. Then by Theorem 1 the powers of P with respect to both circles are equal to <math>PX * PY</math>, and hence by transitive <math>pow(P, \omega_1) = pow(P, \omega_2)</math>. Thus, if point P lies on <math>XY</math>, then the powers of P with respect to both circles are equal. | ||
− | Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on <math>XY</math>. In particular, line <math>PY</math> does not intersect X. Then <math>PY</math> intersects circles <math>\omega_1</math> and <math>\omega_2</math> a second time at distinct points <math>M</math> and <math>N</math>, respectively. (If <math>PY</math> is tangent to <math>\omega_1</math>, for example, we adopt the convention that <math>P = M</math>; similar conventions hold for <math>\omega_2</math>. Power of a Point still holds in this case. Also, notice that <math>M</math> and <math>N</math> cannot both equal <math>P</math>, as <math>PY</math> cannot be tangent to both circles.) Because <math>PM</math> is not equal to <math>PN</math>, <math> | + | Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on <math>XY</math>. In particular, line <math>PY</math> does not intersect X. Then <math>PY</math> intersects circles <math>\omega_1</math> and <math>\omega_2</math> a second time at distinct points <math>M</math> and <math>N</math>, respectively. (If <math>PY</math> is tangent to <math>\omega_1</math>, for example, we adopt the convention that <math>P = M</math>; similar conventions hold for <math>\omega_2</math>. Power of a Point still holds in this case. Also, notice that <math>M</math> and <math>N</math> cannot both equal <math>P</math>, as <math>PY</math> cannot be tangent to both circles.) Because <math>PM</math> is not equal to <math>PN</math>, so <math>PM \cdot PY</math> does not equal <math>PN \cdot PY</math>, and thus by Theorem 1 <math>pow(P, \omega_1)</math> is not congruent to <math>pow(P, \omega_2)</math>, as desired. This completes part (b). |
For the remaining parts, we employ a lemma: | For the remaining parts, we employ a lemma: | ||
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Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.) | Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.) | ||
− | Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.) | + | Now, try to prove '''Theorem 3''' on your own! (Hint: Let P be the intersection of two of the radical axes.)The proof of this theorem along with the proof of '''Theorem 4'''are given in EGMO as '''Example 2.7''' and '''Theorem 2.9'''. |
+ | |||
+ | '''Theorem 5''' | ||
+ | Let <math>AP</math> intersect at <math>BC</math> at <math>M</math> and let <math>(ABP)=\omega_B</math>, <math>(ACP)=\omega_C</math>. Clearly, <math>AP</math> is a radical axis of <math>\omega_B</math>, <math>\omega_C</math>. We see that<cmath>\text{Pow}_{\omega_B}(M)=\text{Pow}_{\omega_C}(M) \implies MB^2=MC^2 \implies MB=MC,</cmath>as desired. | ||
==Exercises== | ==Exercises== | ||
If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. | If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. | ||
− | ''Note:'' No solutions will be provided to the following problems. If you are stuck, ask on the forum. | + | ''Note:'' No solutions will be provided to the following problems(laziness). If you are stuck, ask on the forum. |
− | '''Problem 1.''' Two circles P and Q intersect at X and Y. Point A is located on <math>XY</math> such that AP = 10 and AQ = <s>15</s> 12. If the radius of Q is 7, find the radius of P. <span style="color:#ff0000">Note: An error in this problem previously rendered it unsolvable.</span> | + | '''Problem 1.''' Two circles <math>P</math> and <math>Q</math> intersect at <math>X</math> and <math>Y</math>. Point <math>A</math> is located on <math>XY</math> such that <math>AP = 10</math> and <math>AQ =</math> <s>15</s> <math>12</math>. If the radius of <math>Q</math> is <math>7</math>, find the radius of <math>P</math>. <span style="color:#ff0000">Note: An error in this problem previously rendered it unsolvable.</span> |
'''Problem 2.''' Solve [[2009_USAMO_Problems|2009 USAMO Problem 1]]. If you already know how to solve it. | '''Problem 2.''' Solve [[2009_USAMO_Problems|2009 USAMO Problem 1]]. If you already know how to solve it. | ||
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'''Problem 7.''' Solve [[1995_IMO_Problems/Problem_1|1995 IMO Problem 1]] in two different ways. Compare your solutions with the solutions provided. | '''Problem 7.''' Solve [[1995_IMO_Problems/Problem_1|1995 IMO Problem 1]] in two different ways. Compare your solutions with the solutions provided. | ||
+ | |||
+ | ==Problems== | ||
+ | ===Simple=== | ||
+ | [[File:Point on axes.png|300px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> points <math>D \in BC</math> and <math>E \in BC</math> be given. Denote <math>\Omega = \odot ABC, \omega = \odot AED, G = \omega \cap \Omega \ne A.</math> | ||
+ | Prove that <math>\frac {BF}{FC} = \frac{BE \cdot BD}{CE \cdot CD}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, the order of the points is <math>B,E,F,C,D,</math> as shown on diagram. | ||
+ | <cmath>AF \cdot GF = BF \cdot FC = EF \cdot FD</cmath> | ||
+ | |||
+ | <cmath>BF \cdot FC = (BE + EF) \cdot FC = EF \cdot (FC + CD) \implies BE \cdot FC = EF \cdot CD.</cmath> | ||
+ | <cmath>BF \cdot FC = BF \cdot (EC - EF) = EF \cdot (BD - BF) \implies BF \cdot EC = EF \cdot BD \implies</cmath> | ||
+ | <cmath>\frac {BE \cdot FC}{CD} = EF = \frac {BF \cdot EC}{BD} \implies | ||
+ | \frac {BF}{FC} = \frac{BE \cdot BD}{CE \cdot CD}.</cmath> | ||
+ | |||
+ | ===Intermediate=== | ||
+ | [[2021_AIME_I_Problems#Problem_13|2021 AIME Problem 13]]. Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. | ||
+ | ===Olympiad=== | ||
+ | [[2014_USAMO_Problems#Problem_5|2014 USAMO Problem 5]]. Let <math>ABC</math> be a triangle with orthocenter <math>H</math> and let <math>P</math> be the second intersection of the circumcircle of triangle <math>AHC</math> with the internal bisector of the angle <math>\angle BAC</math>. Let <math>X</math> be the circumcenter of triangle <math>APB</math> and <math>Y</math> the orthocenter of triangle <math>APC</math>. Prove that the length of segment <math>XY</math> is equal to the circumradius of triangle <math>ABC</math>. | ||
+ | |||
+ | [[2017_USAMO_Problems#Problem_3|2017 USAMO Problem 3]]. Let <math>ABC</math> be a scalene triangle with circumcircle <math>\Omega</math> and incenter <math>I.</math> Ray <math>AI</math> meets <math>BC</math> at <math>D</math> and <math>\Omega</math> again at <math>M;</math> the circle with diameter <math>DM</math> cuts <math>\Omega</math> again at <math>K.</math> Lines <math>MK</math> and <math>BC</math> meet at <math>S,</math> and <math>N</math> is the midpoint of <math>IS.</math> The circumcircles of <math>\triangle KID</math> and <math>\triangle MAN</math> intersect at points <math>L_1</math> and <math>L.</math> Prove that <math>\Omega</math> passes through the midpoint of either <math>IL_1</math> or <math>IL.</math> | ||
+ | |||
+ | [[2012_IMO_Problems#Problem_5|2012 IMO Problem 5]]. Let <math>ABC</math> be a triangle with <math>\angle BCA=90^{\circ}</math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let <math>K</math> be the point on the segment <math>AX</math> such that <math>BK=BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL=AC</math>. Let <math>M=\overline{AL}\cap \overline{BK}</math>. Prove that <math>MK=ML</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | [[Category : Geometry]] |
Latest revision as of 08:38, 23 May 2024
Contents
Introduction
The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious USAMO. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success.
Not all theorems will be fully proven in this text. The objective of this document is to introduce you to some key concepts, and then give you a chance to derive some of the beautiful results on your own. In that way, you will understand and retain the information in here much more solidly. Finally, your newfound knowledge will be tested on a few challenging problems that are exemplary examples on how radical axis theory can be used and why it pertains to that situation. I hope after you read this text, you will become a better math student, armed with another tool to solve difficult problems. But, anyway, good luck.
Definitions
The power of point with respect to circle
(with radius
and center
), which shall thereafter be dubbed
, is defined to equal
.
Note that the power of a point is negative if the point is inside the circle.
The radical axis of two non-concentric circles is defined as the locus of the points
such that the power of
with respect to
and
are equal. In other words, if
are the center and radius of
, then a point
is on the radical axis if and only if
(i.e., the radical axis is the line that one gets when you subtract the equations of two circles).
Results
Theorem 1: (Power of a Point)
If a line drawn through point P intersects circle at points A and B, then
.
Theorem 2: (Radical Axis Theorem)
a. The radical axis is a line perpendicular to the line connecting the circles' centers (line ).
b. If the two circles intersect at two common points, their radical axis is the line through these two points.
c. If they intersect at one point, their radical axis is the common internal tangent.
d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to through point A, the unique point on
such that
.
Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center.
Theorem 4: (Radical Centre of Intersecting Circles) (EGMO Theorem 2.9)
Let and
be two circles with centers
and
. Select two points A and B on
and
. Then the following are equivalent
lie on a circle with center
not on line
.
- Lines
and
intersect on the radical axis of
and
.
Theorem 5: (EGMO Lemma 2.11)
Let ABC be a triangle and consider a point in its interior. Suppose that
is tangent to
and
, ray
bisects
Proofs
Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)
Theorem 2 shall be proved here. Assume the circles are and
with centers
and
and radii
and
, respectively. (It may be a good idea for you to draw some circles here.)
First, we tackle part (b). Suppose the circles intersect at points and
and point P lies on
. Then by Theorem 1 the powers of P with respect to both circles are equal to
, and hence by transitive
. Thus, if point P lies on
, then the powers of P with respect to both circles are equal.
Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on . In particular, line
does not intersect X. Then
intersects circles
and
a second time at distinct points
and
, respectively. (If
is tangent to
, for example, we adopt the convention that
; similar conventions hold for
. Power of a Point still holds in this case. Also, notice that
and
cannot both equal
, as
cannot be tangent to both circles.) Because
is not equal to
, so
does not equal
, and thus by Theorem 1
is not congruent to
, as desired. This completes part (b).
For the remaining parts, we employ a lemma:
Lemma 1: Let be a point in the plane, and let
be the foot of the perpendicular from
to
. Then
.
The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.
Lemma 2: There is an unique point P on line such that
.
Proof: First show that P lies between and
via proof by contradiction, by using a bit of inequality theory and the fact that
. Then, use the fact that
(a constant) to prove the lemma.
Lemma 1 shows that every point on the plane can be equivalently mapped to a line on . Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to
, completing part (a).
Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.)
Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.)The proof of this theorem along with the proof of Theorem 4are given in EGMO as Example 2.7 and Theorem 2.9.
Theorem 5
Let intersect at
at
and let
,
. Clearly,
is a radical axis of
,
. We see that
as desired.
Exercises
If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. Note: No solutions will be provided to the following problems(laziness). If you are stuck, ask on the forum.
Problem 1. Two circles and
intersect at
and
. Point
is located on
such that
and
15 . If the radius of
is
, find the radius of
. Note: An error in this problem previously rendered it unsolvable.
Problem 2. Solve 2009 USAMO Problem 1. If you already know how to solve it.
Problem 3. Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of if and only if
.
Problem 4. Solve 2012 USAJMO Problem 1.
Problem 5. Does Theorem 2 apply to circles in which one is contained inside the other? How about internally tangent circles? Concentric circles?
Problem 6. Construct the radical axis of two circles. What happens if one circle encloses the other?
Problem 7. Solve 1995 IMO Problem 1 in two different ways. Compare your solutions with the solutions provided.
Problems
Simple
Let triangle points
and
be given. Denote
Prove that
Proof
WLOG, the order of the points is as shown on diagram.
Intermediate
2021 AIME Problem 13. Circles and
with radii
and
, respectively, intersect at distinct points
and
. A third circle
is externally tangent to both
and
. Suppose line
intersects
at two points
and
such that the measure of minor arc
is
. Find the distance between the centers of
and
.
Olympiad
2014 USAMO Problem 5. Let be a triangle with orthocenter
and let
be the second intersection of the circumcircle of triangle
with the internal bisector of the angle
. Let
be the circumcenter of triangle
and
the orthocenter of triangle
. Prove that the length of segment
is equal to the circumradius of triangle
.
2017 USAMO Problem 3. Let be a scalene triangle with circumcircle
and incenter
Ray
meets
at
and
again at
the circle with diameter
cuts
again at
Lines
and
meet at
and
is the midpoint of
The circumcircles of
and
intersect at points
and
Prove that
passes through the midpoint of either
or
2012 IMO Problem 5. Let be a triangle with
, and let
be the foot of the altitude from
. Let
be a point in the interior of the segment
. Let
be the point on the segment
such that
. Similarly, let
be the point on the segment
such that
. Let
. Prove that
.