Mathematical and Physical Journal
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Problem B. 4054. (January 2008)

B. 4054. The radius of a circle inscribed in a triangle is r. The tangents drawn to the circle parallel to the sides cut three small triangles off the original triangle. Prove that the sum of the radii of the inscribed circles of the small triangles is also r.

(3 pont)

Deadline expired on February 15, 2008.


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

Megoldás: Legyenek a háromszög oldalai a,b,c, területe t. Ha i\in{a,b,c}, akkor az i oldallal párhuzamos érintővel levágott kis háromszög hasonló lesz az eredetihez, ahol a hasonlóság aránya, \alphai=(mi-2r)/mi=1-2r/mi. Ennek alapján a kis háromszögekbe írt körök sugarának összege

(\alpha_a+\alpha_b+\alpha_c)r=\Bigl\{3-2r\Bigl(\frac{1}{m_a}+\frac{1}{m_b}+
\frac{1}{m_c}\Bigr)\Bigr\}r=\Bigl\{3-2r\Bigl(\frac{a}{2t}+\frac{b}{2t}+
\frac{c}{2t}\Bigr)\Bigr\}r=

=\Bigl\{3-2\cdot\frac{r(a+b+c)}{2t}\Bigr\}r=
\Bigl(3-2\cdot\frac{2t}{2t}\Bigr)r=r.


Statistics:

127 students sent a solution.
3 points:97 students.
2 points:28 students.
1 point:1 student.
0 point:1 student.

Problems in Mathematics of KöMaL, January 2008