Mathematical and Physical Journal
for High Schools
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Problem K. 401. (December 2013)

K. 401. In the rectangle ABCD, AB=1 and BC=\sqrt 3. A regular triangle is drawn over side AB on the inside, and another regular triangle on side BC is drawn on the outside. The points P and Q are obtained, respectively. What is the area of triangle APQ?

(6 pont)

Deadline expired on January 10, 2014.

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

Megoldás. Mivel az \(\displaystyle APB\) szabályos háromszög oldala 1, ezért a magassága \(\displaystyle \frac{\sqrt3}{2}\). Ez azt jelenti, hogy \(\displaystyle PQ\) párhuzamos az \(\displaystyle AB\) oldallal, vagyis az \(\displaystyle APQ\) háromszögben az \(\displaystyle A\) csúcshoz tartozó magasság hossza \(\displaystyle \frac{\sqrt3}{2}\). A \(\displaystyle PQ\) hossza \(\displaystyle AB\) felének és a \(\displaystyle CBQ\) háromszög magasságának összegével egyenlő, azaz \(\displaystyle \frac12+\sqrt3\cdot\frac{\sqrt3}{2}=2\). Vagyis a keresett terület: \(\displaystyle T_{APQ}=\frac{2\cdot\frac{\sqrt3}{2}}{2}=\frac{\sqrt3}{2}\).


169 students sent a solution.
6 points:95 students.
5 points:15 students.
4 points:31 students.
3 points:5 students.
2 points:7 students.
1 point:7 students.
0 point:2 students.
Unfair, not evaluated:7 solutions.

Problems in Mathematics of KöMaL, December 2013