Problem K. 196.

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

(6 points)

Deadline expired.

Sorry, the solution is published in Hungarian only.

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 on problem K. 196.

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

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