Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem C. 919. (November 2007)

C. 919. A right-angled triangle is cut into a triangle and a quadrilateral by the perpendicular bisector of its hypotenuse. The ratio of the diagonals of the quadrilateral is \big(1+\sqrt3\,\big):2\sqrt2. Find the acute angles of the right-angled triangle.

(5 pont)

Deadline expired on December 17, 2007.


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

Megoldás:

Mivel E rajta van az AB szakasz felező merőlegesén, ezért EBA\angle=BAC\angle=\alpha.

BDE\triangle\sim ACB\triangle, ezért \frac{BD}{BE}=\frac{b}{c}. A Thalesz-tétel miatt BD=DC, így az átlók aránya \frac{CD}{BE}=\frac{b}{c}<1. Tudjuk, hogy 1+\sqrt3<2\sqrt2, ezért az átlók aránya:

b/c=\sin\alpha=(1+\sqrt3)/2\sqrt2.

Mivel

\sin75^{\circ}=\sin(30^{\circ}+45^{\circ})=\frac12\cdot
\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}=\frac{1+\sqrt3}{2\sqrt2},

ezért

\alpha=75o,

\beta=15o.


Statistics:

274 students sent a solution.
5 points:232 students.
4 points:19 students.
3 points:4 students.
2 points:4 students.
1 point:4 students.
0 point:10 students.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, November 2007