Mathematical and Physical Journal
for High Schools
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Problem C. 864. (September 2006)

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 pont)

Deadline expired on October 16, 2006.

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

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.


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