Problem C. 901. (May 2007)

C. 901. The area of a rectangle ABCD is $100\sqrt5$ . Let P denote the point on side AB, closer to A, that divides AB in a 1:4 ratio. Given that the line segment PD is perpendicular to diagonal AC, calculate the perimeter of the rectangle.

(5 pont)

Deadline expired on June 15, 2007.

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

Megoldás. $APD_{\triangle}\cong BCA_{\triangle}$ , mert szögeik páronként egyenlők. Vagyis $\frac{a}{b}=\frac{b}{\frac{a}{5}}$ , amiből $\left(\frac{a}{b}\right)^2=5$ , vagyis $a=b\sqrt5$ . A terület: $ab=100\sqrt5$ , a helyettesítés után b²=100.

Az oldalak: $a=10\sqrt5$ , b=10.

A kerület: $k=20(1+\sqrt5)$ .

Statistics:

164 students sent a solution.

5 points: 144 students.

4 points: 5 students.

3 points: 2 students.

2 points: 2 students.

1 point: 3 students.

0 point: 1 student.

Unfair, not evaluated: 7 solutionss.

Problems in Mathematics of KöMaL, May 2007

164 students sent a solution.
5 points:	144 students.
4 points:	5 students.
3 points:	2 students.
2 points:	2 students.
1 point:	3 students.
0 point:	1 student.
Unfair, not evaluated:	7 solutionss.

Already signed up?
New to KöMaL?