Mathematical and Physical Journal
for High Schools
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Problem K. 383. (September 2013)

K. 383. Base AB of the regular triangle ABC is extended beyond vertex A by two fifths of the length AB, to get point P. Point P is connected to the point Q that lies on side AC (closer to A) and divides it 2:3. The resulting line intersects line CB at point R. Given that AP=2684, find the length of CR.

(6 pont)

Deadline expired on October 10, 2013.

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

Megoldás. \(\displaystyle PA=QA=\frac25\cdot AB=2684\), amiből egyrészt \(\displaystyle AB=6710\), másrészt \(\displaystyle APQ\angle=AQP\angle=\frac{180^{\circ}-60^{\circ}}{2}=30^{\circ}\) következik. Mivel \(\displaystyle RQC\angle=AQP\angle=30^{\circ}\) és \(\displaystyle QCR\angle=ACB\angle=60^{\circ}\), ezért \(\displaystyle CRQ\angle=180^{\circ}-(RQC\angle+QCR\angle)=90^{\circ}\) és a \(\displaystyle CRQ\) háromszög egy szabályos háromszög fele. Ebből következik, hogy \(\displaystyle CR=QC/2=(AC-AQ)/2=(6710-2684)/2=4026/2=2013\).


241 students sent a solution.
6 points:153 students.
5 points:34 students.
4 points:22 students.
3 points:6 students.
2 points:5 students.
1 point:10 students.
0 point:9 students.
Unfair, not evaluated:2 solutionss.

Problems in Mathematics of KöMaL, September 2013