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C. 864. A triangle is drawn on squared paper. The lengths of the sides are 2\sqrt{10}, 3\sqrt5 and 5 units. Prove that the measure of the smallest angle is 45o.

(Suggested by G. Holló, Budapest)

(5 points)

Deadline expired on 16 October 2006.


Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás: Mivel az 5 a háromszög legkisebb oldala, ezért úgy írjuk fel a koszinusz-tételt, hogy az 5-tel szemközti szög szerepjen benne:

5^2=(2\sqrt{10})^2 +(3\sqrt5)^2-2\cdot(2\sqrt{10})\cdot(3\sqrt5)\cdot\cos\varphi.

Ebből \cos\varphi=\frac{60}{60\sqrt2}=\frac{1}{\sqrt2}, vagyis \varphi=45o.


Statistics on problem C. 864.
566 students sent a solution.
5 points:402 students.
4 points:104 students.
3 points:8 students.
2 points:16 students.
0 point:30 students.
Unfair, not evaluated:6 solutions.


  • Problems in Mathematics of KöMaL, September 2006

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