Right-angled triangle and incircles

[ins-]

Let $ABC$ is a right-angled triangle ($\angle{C}=90^o$) with incircle $k$. $l_1$ is a line tangent to $k$ and parallel to $BC$ intersecting $AC$ and $AB$ at the points $K$ and $L$. $l_2$ is a line tangent to $k$ and parallel to $AC$ intersecting $BC$ and $AB$ at the points $M$ and $N$. $l_3$ is a line tangent to $k$ and parallel to $AB$ intersecting $AC$ and $BC$ at the points $P$ and $Q$. If $r_1$, $r_2$ and $r_3$ are the inradii of the triangles $AKL$, $BMN$ and $CPQ$ - express $r_3$ in terms of $r_1$ and $r_2$.

(Оффтоп)

I composed this problem and know the solution. I'm interested to see different ideas.

[DeBill]

ins-
$r_1r_2=r_3\cdot (r_1+r_2+r_3)$
Решение: грубое с элементами извращения тригонометрическое насилие....

[ins-]

Wow! It deserves to be seen.I have two very elegant solutions without trigonometry. For the first one I saw an idea from a guy, named Peter Scales. It inspired me to compose the problem and another, probably Russian, problem I saw when I was 16 years old. If you like - I can show the solutions. I suppose even more approaches are possible. The inradius also can be found.

[DeBill]

ins- в сообщении #1106263 писал(а):

If you like

Yes

[ins-]

http://artofproblemsolving.com/communit ... 61p6001194 the first one is completed by a friend of mine and the second is just sketched in post 6 by me. Excuse me for redirecting you. My goal was to create a not very easy, and not very hard and interesting problem considering right-angled triangles, but it seems - it is a little harder than I expected.

[DeBill]

ins-
Красиво!
Конечно, я находил радиусы из подобия. Но, для прямоугольного тр-ка, слишком много хороших соотношений...
В конце концов, я плюнул на попытки сделать "красиво", и добил все тригонометрией...

[ins-]

I'm interested to see different approaches. From every solution a person can learn something. If you like - you can post your solution. Believe or not I cannot use trigonometry in a meaningful way for this problem. I managed to find dependencies in three more similar configurations involving right-angled triangles between the radiuses of three inscribed circles, but used Fukagawa and Pedoe's book dedicated to Sangaku. For them there are more calculations and they are not so good looking. It was the reason to share this one.

[Sergic Primazon]

ins- в сообщении #1106147 писал(а):

Let $ABC$ is a right-angled triangle ($\angle{C}=90^o$) with incircle $k$. $l_1$ is a line tangent to $k$ and parallel to $BC$ intersecting $AC$ and $AB$ at the points $K$ and $L$. $l_2$ is a line tangent to $k$ and parallel to $AC$ intersecting $BC$ and $AB$ at the points $M$ and $N$. $l_3$ is a line tangent to $k$ and parallel to $AB$ intersecting $AC$ and $BC$ at the points $P$ and $Q$. If $r_1$, $r_2$ and $r_3$ are the inradii of the triangles $AKL$, $BMN$ and $CPQ$ - express $r_3$ in terms of $r_1$ and $r_2$.

(Оффтоп)

I composed this problem and know the solution. I'm interested to see different ideas.

$$KX=YL=LZ=UQ=r_1, VM=NW=ZN=PU=r_2$$

Поэтому :

$$(r_1+r_3)^2+(r_2+r_3)^2=(r_1+r_2)^2$$

[ins-]

It is the 5-th or sixth approach for this problem to be solved. I wonder how you managed to find such a construction. Very beautiful! Thanks for sharing it! I have only one question - how did you get KX=YL?

[DeBill]

Блеск!
ins-
Это - известное свойство вписанной и вневписанной окружностей. Полезно и формулы помнить для этих отрезков - через длины сторон.

[ins-]

:) I understood it. I didn't need excircles' properties to solve the problem, but it is a good occasion to learn something new.

[Red_Herring]

Слава богу, первый чертёж появился.

[mihiv]

ins- в сообщении #1106263 писал(а):

The inradius also can be found.

Если $r$- радиус вписанной окружности большого треугольника, то $r=r_1+r_2+r_3.$
Это справедливо для любого треугольника, не только прямоугольного.

[ins-]

The configuration I started from was for any triangle. I saw for it $r=r_1+r_2+r_3$ ( it was a problem in an old Bulgarian book ). Then I started to ask myself questions "what if" and created the initial problem. I wondered if to not mention $r_3$ at all in the problem (it is a combination of 2-3 problems) and to ask to find a dependency between $r$, $r_1$ and r_2, but decided to put the accent on the fact that r_3 can be expressed in terms of r_1 and r_2.

[mihiv]

Обозначим $S$- площадь треугольника $ABC,$ пусть $ a$- сторона, противолежащая углу $A$, аналогично обозначим стороны $b$ и $c$.
Каждая касательная к вписанной в тр-к $ABC$ окружности делит его на меньший тр-к, подобный тр-ку $ABC$, и прямоугольную трапецию.
Площадь этой трапеции запишем, один раз,используя подобие треугольников, а другой- непосредственно. В результате получим (для треугольника с радиусом $r_1): S(1-\frac {r_1^2}{r^2})=ra(1+\frac {r_1}{r})$ или $a=\frac Sr(1-\frac {r_1}{r})$. Таким же образом $b=\frac Sr(1-\frac {r_2}{r}), c=\frac Sr(1-\frac {r_3}{r})$. По теореме Пифагора $a^2+b^2=c^2.$ Отсюда, учитывая, что $r=r_1+r_2+r_3$: $(r_1+r_3)^2+(r_2+r_3)^2=(r_1+r_2)^2$.