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 45^o.

(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 $\varphi$ =45^o.

Statistics:

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 solutionss.

Problems in Mathematics of KöMaL, September 2006

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 solutionss.

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