# Difference between revisions of "1995 IMO Problems/Problem 1"

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Let <math>A,B,C,D</math> be four distinct points on a line, in that order. The circles with diameters <math>AC</math> and <math>BD</math> intersect at <math>X</math> and <math>Y</math>. The line <math>XY</math> meets <math>BC</math> at <math>Z</math>. Let <math>P</math> be a point on the line <math>XY</math> other than <math>Z</math>. The line <math>CP</math> intersects the circle with diameter <math>AC</math> at <math>C</math> and <math>M</math>, and the line <math>BP</math> intersects the circle with diameter <math>BD</math> at <math>B</math> and <math>N</math>. Prove that the lines <math>AM,DN,XY</math> are concurrent. | Let <math>A,B,C,D</math> be four distinct points on a line, in that order. The circles with diameters <math>AC</math> and <math>BD</math> intersect at <math>X</math> and <math>Y</math>. The line <math>XY</math> meets <math>BC</math> at <math>Z</math>. Let <math>P</math> be a point on the line <math>XY</math> other than <math>Z</math>. The line <math>CP</math> intersects the circle with diameter <math>AC</math> at <math>C</math> and <math>M</math>, and the line <math>BP</math> intersects the circle with diameter <math>BD</math> at <math>B</math> and <math>N</math>. Prove that the lines <math>AM,DN,XY</math> are concurrent. | ||

+ | |||

+ | == Hint == | ||

+ | Think about Radical Axis, Power of a Point and Radical Center. | ||

== Solution == | == Solution == | ||

− | Since <math>M</math> is on the circle with diameter <math>AC</math>, we have <math>\angle AMC=90</math> and so <math>\angle MCA=90-A</math>. We | + | Since <math>M</math> is on the circle with diameter <math>AC</math>, we have <math>\angle AMC=90</math> and so <math>\angle MCA=90-A</math>. We similarly find that <math>\angle BND=90</math>. Also, notice that the line <math>XY</math> is the radical axis of the two circles with diameters <math>AC</math> and <math>BD</math>. Thus, since <math>P</math> is on <math>XY</math>, we have <math>PN\cdot PB=PM\cdot PC</math> and so by the converse of Power of a Point, the quadrilateral <math>MNBC</math> is cyclic. Thus, <math>90-A=\angle MCA=\angle BNM</math>. Thus, <math>\angle MND=180-A</math> and so quadrilateral <math>AMND</math> is cyclic. Let the name of the circle <math>AMND</math> be <math>O</math> . Then, the radical axis of <math>O</math> and the circle with diameter <math>AC</math> is line <math>AM</math>. Also, the radical axis of <math>O</math> and the circle with diameter <math>BD</math> is line <math>DN</math>. Since the pairwise radical axes of 3 circles are concurrent, we have <math>AM,DN,XY</math> are concurrent as desired. |

+ | |||

+ | ==Solution 2== | ||

+ | Let <math>AM</math> and <math>PT</math> (a subsegment of <math>XY</math>) intersect at <math>Z</math>. Now, assume that <math>Z, N, P</math> are not collinear. In that case, let <math>ZD</math> intersect the circle with diameter <math>BD</math> at <math>N'</math> and the circle through <math>D, P, T</math> at <math>N''</math>. | ||

+ | |||

+ | We know that <math>\angle AMC = \angle BND = \angle ATP = 90^\circ</math> via standard formulae, so quadrilaterals <math>AMPT</math> and <math>DNPT</math> are cyclic. Thus, <math>N'</math> and <math>N''</math> are distinct, as none of them is <math>N</math>. Hence, by Power of a Point, <cmath>ZM * ZA = ZP * ZT = ZN'' * ZD.</cmath> However, because <math>Z</math> lies on radical axis <math>TP</math> of the two circles, we have <cmath>ZM * ZA = ZN' * ZD.</cmath> Hence, <math>ZN'' = ZN'</math>, a contradiction since <math>D</math> and <math>D'</math> are distinct. We therefore conclude that <math>Z, N, D</math> are collinear, which gives the concurrency of <math>AM, XY</math>, and <math>DN</math>. This completes the problem. | ||

+ | |||

+ | == Solution 3== | ||

+ | Let <math>AM</math> and <math>XY</math> intersect at <math>Z</math>. Because <math>\angle AMC = \angle BND = \angle APT = 90^\circ</math>, we have quadrilaterals <math>AMPT</math> and <math>DNPT</math> cyclic. Therefore, <math>Z</math> lies on the radical axis of the two circumcircles of these quadrilaterals. But <math>Z</math> also lies on radical axis <math>XY</math> of the original two circles, so the power of <math>Z</math> with respect to each of the four circles is all equal to <math>ZM * ZA</math>. Hence, <math>Z</math> lies on the radical axis <math>DN</math> of the two circles passing through <math>D</math> and <math>N</math>, as desired. | ||

+ | |||

+ | what is <math>T</math> here? | ||

+ | |||

+ | ==Discussion== | ||

+ | ''Lemma:'' The radical axis of two pairs of circles <math>O_1</math>, <math>O_2</math> and <math>O_3</math>, <math>O_4</math> are the same line <math>l</math>. Furthermore, <math>O_1</math> and <math>O_2</math> intersect at <math>A</math> and <math>B</math>, and <math>O_3</math> and <math>O_4</math> intersect at <math>C</math> and <math>D</math>. Then <math>A, B, C,</math> and <math>D</math> are concyclic. | ||

+ | |||

+ | The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point. | ||

+ | |||

+ | Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles. | ||

+ | |||

+ | Solution 1 also gives a trivial lemma that can also be useful: | ||

+ | |||

+ | ''Lemma 2:'' Chords <math>AB</math> of <math>\omega_1</math> and <math>CD</math> of <math>\omega_2</math> intersect on the segment <math>XY</math> formed from the intersections of the two circles. Then <math>A, B, C, D</math> are concyclic. | ||

+ | |||

+ | Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL! | ||

==See also== | ==See also== | ||

[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |

## Latest revision as of 19:03, 12 October 2021

## Problem

Let be four distinct points on a line, in that order. The circles with diameters and intersect at and . The line meets at . Let be a point on the line other than . The line intersects the circle with diameter at and , and the line intersects the circle with diameter at and . Prove that the lines are concurrent.

## Hint

Think about Radical Axis, Power of a Point and Radical Center.

## Solution

Since is on the circle with diameter , we have and so . We similarly find that . Also, notice that the line is the radical axis of the two circles with diameters and . Thus, since is on , we have and so by the converse of Power of a Point, the quadrilateral is cyclic. Thus, . Thus, and so quadrilateral is cyclic. Let the name of the circle be . Then, the radical axis of and the circle with diameter is line . Also, the radical axis of and the circle with diameter is line . Since the pairwise radical axes of 3 circles are concurrent, we have are concurrent as desired.

## Solution 2

Let and (a subsegment of ) intersect at . Now, assume that are not collinear. In that case, let intersect the circle with diameter at and the circle through at .

We know that via standard formulae, so quadrilaterals and are cyclic. Thus, and are distinct, as none of them is . Hence, by Power of a Point, However, because lies on radical axis of the two circles, we have Hence, , a contradiction since and are distinct. We therefore conclude that are collinear, which gives the concurrency of , and . This completes the problem.

## Solution 3

Let and intersect at . Because , we have quadrilaterals and cyclic. Therefore, lies on the radical axis of the two circumcircles of these quadrilaterals. But also lies on radical axis of the original two circles, so the power of with respect to each of the four circles is all equal to . Hence, lies on the radical axis of the two circles passing through and , as desired.

what is here?

## Discussion

*Lemma:* The radical axis of two pairs of circles , and , are the same line . Furthermore, and intersect at and , and and intersect at and . Then and are concyclic.

The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.

Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.

Solution 1 also gives a trivial lemma that can also be useful:

*Lemma 2:* Chords of and of intersect on the segment formed from the intersections of the two circles. Then are concyclic.

Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!