Mathematical and Physical Journal
for High Schools
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Problem B. 4203. (October 2009)

B. 4203. A common tangent of two intersecting circles touches them at the points A and B, and the line segment connecting their centres intersects them at C and D, respectively. Prove that ABCD is a cyclic quadrilateral.

(4 pont)

Deadline expired on November 10, 2009.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A körök középpontját jelölje \(\displaystyle M\), illetve \(\displaystyle N\) az ábrán látható módon. Mivel \(\displaystyle BAC\sphericalangle+CAM\sphericalangle=90^\circ\), az \(\displaystyle AMC\) egyenlőszárú háromszögben \(\displaystyle AMC\sphericalangle= 180^\circ -2CAM\sphericalangle=2BAC\sphericalangle\). Hasonlóképpen, \(\displaystyle BND\sphericalangle=2ABC\sphericalangle\). Az \(\displaystyle AM\) és \(\displaystyle BN\) szakaszok párhuzamossága miatt \(\displaystyle AMC\sphericalangle+BND\sphericalangle=180^\circ\), ahonnan \(\displaystyle BAC\sphericalangle+ABC\sphericalangle=90^\circ\), vagyis az \(\displaystyle ABC\) háromszög a \(\displaystyle C\) csúcsnál derékszögű. Hasonlóképpen \(\displaystyle ADB\sphericalangle=90^\circ\) is igaz. A \(\displaystyle C\) és \(\displaystyle D\) pontok tehát az \(\displaystyle AB\) szakasz fölé emelt Thalesz-körön vannak, az \(\displaystyle ABCD\) négyszög valóban húrnégyszög.


Statistics:

144 students sent a solution.
4 points:97 students.
3 points:13 students.
2 points:5 students.
1 point:5 students.
0 point:17 students.
Unfair, not evaluated:7 solutionss.

Problems in Mathematics of KöMaL, October 2009