Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem B. 3844. (October 2005)

B. 3844. The legs of a right-angled triangle are a and b, its hypotenuse is c. The radius of the escribed circle drawn to the leg b (i.e. the one touching that leg on the outside and also touching the extensions of the other two sides) is b. Prove that b+c=a+2b.

(3 pont)

Deadline expired on November 15, 2005.

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

Az ábra jelöléseivel élve OX=OY=OZ=a. Az OXCY négyszögnek három szöge is derékszög, ezért az négyzet, vagyis CX=CY=a. Az egybevágó OXB és OZB derékszögű háromszögekben BZ=BX=a-a. AY=AZ miatt

2AY=AY+AZ=(b+a)+(c+a-a)=a+b+c,

ahonnan 2(b+a)=a+b+c, vagyis b+2a=a+c.

### Statistics:

 313 students sent a solution. 3 points: 211 students. 2 points: 83 students. 1 point: 10 students. 0 point: 7 students. Unfair, not evaluated: 2 solutions.

Problems in Mathematics of KöMaL, October 2005