Problem B. 5303. (March 2023)
B. 5303. The isosceles right-angled triangle \(\displaystyle ABC\) has its right angle at \(\displaystyle C\). \(\displaystyle D\) is an interior point of side \(\displaystyle BC\) such that the angle \(\displaystyle CDA\) is \(\displaystyle 75^\circ\). Given that triangle \(\displaystyle ADC\) has unit area, prove that \(\displaystyle BD = 2\).
Proposed by M. Hujter, Budapest
(4 pont)
Deadline expired on April 11, 2023.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. A \(\displaystyle DAB\) szög felezője messe a \(\displaystyle BC\) oldalt \(\displaystyle E\)-ben, legyen továbbá \(\displaystyle AC=BC=m\).
Nyilván \(\displaystyle CAD\sphericalangle = 15^{\circ}\), így \(\displaystyle DAE\sphericalangle = EAB\sphericalangle = 15^{\circ}\). A \(\displaystyle CAE\) derékszögű háromszög \(\displaystyle A\)-nál levő szöge \(\displaystyle 30^{\circ}\) lévén \(\displaystyle AE=m\cdot \dfrac{2}{\sqrt{3}}\) és \(\displaystyle CE=AE/2= \dfrac{m}{\sqrt{3}}\). Mivel \(\displaystyle AD\) felezi a \(\displaystyle CAE\) szöget, a szögfelező tétel szerint \(\displaystyle \dfrac{CD}{DE}=\dfrac{m}{m\cdot \dfrac{2}{\sqrt{3}}} = \dfrac{\sqrt{3}}{2}\). Így
\(\displaystyle DE = CE - CD = \dfrac{m}{\sqrt{3}} - CD,\)
\(\displaystyle \dfrac{\sqrt{3}}{2} = \dfrac{CD}{DE} = \dfrac{CD}{\dfrac{m}{\sqrt{3}} - CD},\)
amiből \(\displaystyle CD=\dfrac{m}{2+\sqrt{3}}\).
A megfelelő háromszögek területére
\(\displaystyle \dfrac{t_{ACD}}{t_{ABC}} = \dfrac{CD}{m} = \dfrac{1}{2+\sqrt{3}}, \)
\(\displaystyle 1=t_{ACD} = \dfrac{1}{2+\sqrt{3}} \cdot \dfrac{1}{2}\cdot m^2, \)
innen pedig \(\displaystyle m^2 = 4+2\sqrt{3}=(1+\sqrt{3})^2\), \(\displaystyle m=1+\sqrt{3}\). Végül
\(\displaystyle BD = m - CD = 1+\sqrt{3} - \dfrac{m}{2+\sqrt{3}} = 1+\sqrt{3} - \dfrac{1+\sqrt{3}}{2+\sqrt{3}} = (1+\sqrt{3})\cdot \big(1- \dfrac{1}{2+\sqrt{3}} \big) = (1+\sqrt{3})\cdot \dfrac{1+\sqrt{3}}{2+\sqrt{3}} = 2. \)
Statistics:
130 students sent a solution. 4 points: 117 students. 3 points: 4 students. 0 point: 1 student. Unfair, not evaluated: 6 solutionss.
Problems in Mathematics of KöMaL, March 2023