Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem B. 4891. (September 2017)

B. 4891. The circles \(\displaystyle S_1\), \(\displaystyle S_2\), \(\displaystyle S_3\) pairwise touch each other on the outside. Let \(\displaystyle A\), \(\displaystyle B\) and \(\displaystyle C\) denote the common points of the circles \(\displaystyle S_1\) and \(\displaystyle S_2\), \(\displaystyle S_1\) and \(\displaystyle S_3\), \(\displaystyle S_2\) and \(\displaystyle S_3\), respectively. Line \(\displaystyle AB\) intersects the circles \(\displaystyle S_2\) and \(\displaystyle S_3\) again at points \(\displaystyle D\) and \(\displaystyle E\), respectively. Line \(\displaystyle DC\) intersects circle \(\displaystyle S_3\) again at \(\displaystyle F\). Prove that triangle \(\displaystyle DEF\) is right-angled.

(Kvant)

(5 pont)

Deadline expired on October 10, 2017.


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

Megoldásvázlat. Azt fogjuk belátni, hogy \(\displaystyle DEF\sphericalangle =90^{\circ}\). A \(\displaystyle B, E, F\) és \(\displaystyle C\) pontok az \(\displaystyle S_{3}\) kör pontjai, a \(\displaystyle BEFC\) húrnégyszög, tehát elegendő belátni, hogy a \(\displaystyle BCD\sphericalangle\) derékszög.

Legyen az \(\displaystyle S_1\), \(\displaystyle S_2\) és \(\displaystyle S_3\) körök középpontja \(\displaystyle O_1\), \(\displaystyle O_2\) és \(\displaystyle O_3\).

Az \(\displaystyle A\), \(\displaystyle B\) és \(\displaystyle C\) érintési pontok, tehát rajta vannak rendre az \(\displaystyle O_1O_2\), \(\displaystyle O_1O_3\) és \(\displaystyle O_2O_3\) szakaszokon. \(\displaystyle AO_1=BO_1\), \(\displaystyle BO_3=CO_3\) és \(\displaystyle CO_2=AO_2\), ami azt jelenti, hogy \(\displaystyle A\), \(\displaystyle B\) és \(\displaystyle C\) egyben az \(\displaystyle O_1O_2O_3\) háromszög beírt körének érintési pontjai. A beírt kör középpontja legyen \(\displaystyle K\). Az \(\displaystyle O_1O_2O_3\) háromszög szögei pedig \(\displaystyle \alpha\), \(\displaystyle \beta\) és \(\displaystyle \gamma\). Az egyenlő szárú háromszögekből azonnal számolhatóak az \(\displaystyle ABC\) háromszög szögei is: \(\displaystyle ABC\sphericalangle=\frac{\alpha+\gamma}{2}\), \(\displaystyle BCA\sphericalangle=\frac{\beta+\gamma}{2}\), \(\displaystyle CAB\sphericalangle=\frac{\beta+\alpha}{2}\)

Most alkalmazzuk az \(\displaystyle S_2\) körben a kerületi-középponti szögek tételét:

\(\displaystyle ADC\sphericalangle=\frac{AO_2C\sphericalangle}{2}=\frac{\beta}{2}.\)

\(\displaystyle DCB\sphericalangle=180°-ADC\sphericalangle-ABC\sphericalangle=180°-\frac{\alpha+\beta+\gamma}{2}=90°,\)

ezért \(\displaystyle DEF\sphericalangle\) is derékszög.


Statistics:

113 students sent a solution.
5 points:90 students.
4 points:7 students.
3 points:2 students.
2 points:8 students.
1 point:6 students.

Problems in Mathematics of KöMaL, September 2017