Mathematical and Physical Journal
for High Schools
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Problem C. 1108. (January 2012)

C. 1108. In a right-angled triangle ABC, the altitude drawn to the hypotenuse is CD. Prove that the sum of the areas of the circles inscribed in triangles ADC and BCD equals the area of the inscribed circle of triangle ABC.

(5 pont)

Deadline expired on February 10, 2012.


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

Megoldás. Az \(\displaystyle ACD\) háromszög beírt körének sugarát jelölje \(\displaystyle r_A\), a \(\displaystyle BCD\) háromszög beírt körének sugarát \(\displaystyle r_B\), az \(\displaystyle ABC\) háromszögét pedig \(\displaystyle r\). Az \(\displaystyle ABC\) háromszöghöz hasonló mind az \(\displaystyle ADC\), mind a \(\displaystyle BCD\) háromszög: a hasonlóságok arányát felírva rendre kapjuk, hogy \(\displaystyle \frac{r_B}{r}=\frac ac\) és \(\displaystyle \frac{r_A}{r}=\frac bc\). A kisebb körök területének összege

\(\displaystyle r_A^2 \pi + r_B^2 \pi = \frac{a^2 r^2 + b^2 r^2}{c^2}\cdot \pi=r^2 \pi,\)

ahol felhasználtuk Pithagorasz tételét.


Statistics:

228 students sent a solution.
5 points:135 students.
4 points:47 students.
3 points:20 students.
2 points:11 students.
1 point:4 students.
0 point:4 students.
Unfair, not evaluated:7 solutions.

Problems in Mathematics of KöMaL, January 2012