Difference between revisions of "1995 IMO Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
− | 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>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><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, | + | 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, XY</math>, and <math>DN</math>. This completes the problem. |
== Solution 3== | == Solution 3== |
Revision as of 18:19, 28 August 2014
Contents
[hide]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
Radical axis and radical center! Think RADICAL-ly!
Solution
Since is on the circle with diameter , we have and so . We simlarly find that . Also, notice that the line is the radical axis of the two circles with diameters and . Thus, since is on , we have $PN\cdotPB=PM\cdot PC$ (Error compiling LaTeX. Unknown error_msg) and so by the converse of Power of a Point, the quadrilateral is cyclic. Thus, . Thus, and so quadrilateral is cyclic. Let the circle which contains the points be cirle . 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, so . But as a result lies on the radical axis of the two original circles (with diameters and ), so lies on as well.