Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem K. 196. (January 2009)

K. 196. In a kite, there is a 90o angle opposite the 60o angle. It is symmetrical to its diagonal of length 10 cm. Find its perimeter.

(6 pont)

Deadline expired on February 10, 2009.


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

Megoldás. Az ABC szabályos háromszögben: BE=\frac{\sqrt3}{2}a. Az ACD egyenlő szárú derékszögű háromszögben: DE=\frac a2. Mivel BD=10, ezért \frac{\sqrt3}{2}a+\frac a2=10 cm. Ebből kapjuk: a=\frac{20}{\sqrt3+1}.

Az AED egyenlő szárú derékszögű háromszögben: AD=b=\frac a2\sqrt2=\frac{10}{\sqrt3+1}\sqrt2.

k=2(a+b)=2\left(\frac{20}{\sqrt3+1}
+\frac{10}{\sqrt3+1}\cdot\sqrt2
\right)=\frac{40+20\sqrt2}{\sqrt3+1}\approx 24,99.

Vagyis a deltoid kerülete kb. 25 cm.


Statistics:

159 students sent a solution.
6 points:109 students.
5 points:8 students.
4 points:9 students.
3 points:2 students.
2 points:2 students.
1 point:3 students.
0 point:13 students.
Unfair, not evaluated:13 solutions.

Problems in Mathematics of KöMaL, January 2009