Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 IMO Problems/Problem 1"
m (→‎Solution)
 
(12 intermediate revisions by 8 users not shown)
Line 4: Line 4:
  
 
== Hint ==
 
== Hint ==
−
Radical axis and radical center! Think RADICAL-ly!
+
Think about Radical Axis, Power of a Point and Radical Center.
  
 +
== Solution 1==
 +
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>.
  
−
== Solution ==
+
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.
−
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 simlarly 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\cdotPB=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 circle which contains the points <math>AMND</math> be cirle <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==
+
== Solution 3==
−
Let <math>AM</math> and <math>PT</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>.
+
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:
  
−
We know that <math><AMC = <BND = <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, PT</math>, and <math>DN</math>. This completes the problem.
+
''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.
  
−
== Solution 3==
+
Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!
−
Let <math>AM</math> and <math>DN</math> intersect at <math>Z</math>. Because <math><AMC = <BND = <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, so <math>ZM * ZA = ZN * ZD</math>. But as a result <math>Z</math> lies on the radical axis of the two original circles (with diameters <math>AC</math> and <math>BD</math>), so <math>Z</math> lies on <math>TP</math> as well.
 
  
 
==See also==
 
==See also==
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 +
{{IMO box|year=1995|before=First Question|num-a=2}}

Latest revision as of 04:48, 4 September 2024

Problem

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


Hint

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

Solution 1

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

Solution 2

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

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

Solution 3

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

what is $T$ here?

Discussion

Lemma: The radical axis of two pairs of circles $O_1$, $O_2$ and $O_3$, $O_4$ are the same line $l$. Furthermore, $O_1$ and $O_2$ intersect at $A$ and $B$, and $O_3$ and $O_4$ intersect at $C$ and $D$. Then $A, B, C,$ and $D$ 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 $AB$ of $\omega_1$ and $CD$ of $\omega_2$ intersect on the segment $XY$ formed from the intersections of the two circles. Then $A, B, C, D$ are concyclic.

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

See also

1995 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_1&oldid=226432"