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What's the area of a right triangle if the quotient of its legs is $1.05$ and the difference between the radii of the inscribed and circumscribed circles is $17$?

I've been trying to solve this and I've got: ($R$ - radius of circumscribed circle, $r$ - radius of inscribed circle)

$1.$ $ \frac{a}{b}=1.05$

$2.$ $c^2=a^2+b^2$

$3.$ $a + b - 2r = c$

$4.$ $c-2r=34$

$5.$ $ab=(a+b+c)r$

Using the first four equations, I can substitute for one of the legs from $1.$ and for $r$ through $4.$ which leaves me with

$b(2.05)-2c=34$

$c=b\sqrt{1.05^2+1}$

However, solving this simply evades me, as I don't find myself getting rid of the square root which I don't know how to calculate.

I do know my equations give the right answer so I'm probably missing a simpler way to solve the system of equations.

Help much appreciated.

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    $\begingroup$ I got the following quadratic in $b$ after simplification: $4.2075b^2 + 139.4b - 1156 = 0$ which factorises as $(b+40)(4.2075b - \frac{1156}{40})$. $\endgroup$ – Shraddheya Shendre Apr 6 '17 at 20:03
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You are missing that $2R=c$, so

$$(2R)^2=a^2+b^2=(1+1.05^2)b^2$$ but $R=17+r$, so

$$4(17+r)^2=(1+1.05^2)b^2 \quad (1)$$

We also have: $$a+b-2r=2R\to(1+1.05)b=2(r+R)=2(17+2r)\quad (2)$$

Combining $(1)$ and $(2)$ you get:

$$\frac{b\sqrt{1+1.05^2}}{2}-17=\frac{1}{2}\left(\frac{b(1+1.05)}{2}-17\right)$$

$$\frac{1.45b}{2}-17=\frac{1}{2}\left(1.025b-17\right)\to1.45b-34=1.025b-17\\ 0,425b=17\to b=40$$

Now you can calculate $a=40\cdot 1.05=42$.

The area will be $$A=\frac{40\cdot 42}{2}=840$$.

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  • $\begingroup$ I don't think he missed $c = 2R$, he has (correctly) written $2R - 2r = 34$ as $c - 2r = 34$. In fact his system of equations is complete. He is apparently having trouble solving it. $\endgroup$ – Shraddheya Shendre Apr 6 '17 at 20:06
  • $\begingroup$ @ShraddheyaShendre: right! Did you understand the solution? $\endgroup$ – Arnaldo Apr 6 '17 at 20:07
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    $\begingroup$ Yes, see my comment underneath the question. $\endgroup$ – Shraddheya Shendre Apr 6 '17 at 20:07
  • $\begingroup$ Hmmm, I have missed that there is a nice root of $2.1025$. This makes sense. I assume this is the simplest way to do this problem? $\endgroup$ – John Doe Apr 6 '17 at 20:19
  • $\begingroup$ @JohnDoe: You were very close. In fact it is simple. I just don't know if it the simplest. $\endgroup$ – Arnaldo Apr 6 '17 at 20:21
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You have correctly arrived at the $2$ equations, namely $$2.05b - 2c = 34$$ and $$c = b \sqrt{1 + 1.05^{2}}$$ Substitute $c$ from second equation into first equation to get $$2.05b - 34 = 2c = 2b\sqrt{1 + 1.05^{2}}$$ Now square both the sides and simplify to get the following quadratic $$4.2075b^2 + 139.4b - 1156 = 0 \implies (b+40)(4.2075b - \frac{1156}{40}) = 0$$ Taking the positive root of $b$ gives $b = 6.\overline{86}$
Now the area could be calculated as $A = \frac{ab}{2} = 0.525 b^2$

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The inradius of a right triangle in terms of its sides is given by the relationship $$r \frac{a+b+c}{2} = rs =|\triangle ABC| = \frac{ab}{2},$$ hence $$r = \frac{ab}{a+b+c}.$$ The circumradius is trivially $$R = \frac{c}{2}.$$ Thus the given conditions may be summarized as $$17 = R - r = \frac{c}{2} - \frac{ab}{a+b+c} = \frac{(a+b)c + c^2 - 2ab}{2(a+b+c)} = \frac{(a+b)c + (a-b)^2}{2(a+b+c)}, \\ \frac{a}{b} = \frac{21}{20}, \\ a^2 + b^2 = c^2.$$ We note that because $21^2 + 20^2 = 29^2$, we have $$a : b : c \equiv 21 : 20 : 29,$$ thus the computation is greatly simplified by letting $a = 21k$, $b = 20k$, $c = 29k$, to obtain from the first equation $$17(2)(70)k = (41)(29)k^2 + (21-20)^2 k^2,$$ or $$2380k = 1190k^2,$$ or $k = 2$. Thus the desired triangle is $(a,b,c) = (42, 40, 58)$ and its area is $$|\triangle ABC| = 840.$$

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Here is a rather simple solution using analytic geometry.

enter image description here

Let us call $AOB$ the right triangle, taking the main vertex $O$ as the origin, and axes resp. directed by OA and OB.

By assumption, the coordinates of $A$ and $B$ are resp. $(21s,0)$ and $(0,20s)$ for a certain $s$. Hypotenuse (H) has thus length $\sqrt{(20s)^2+(21s)^2}=29s$. Therefore the radius of the circumscribed circle is the half of the hypotenuse: $R:=\tfrac{29}{2}s$.

The incenter (center of the inscribed circle), being on the angle bissector of right angle $A$, has coordinates $(t,t)$, where $t$ is the radius of the incircle.

The constraint on the radii becomes:

$$\tag{1}\tfrac{29}{2}s-t=17$$

The equation of line AB is $\tfrac{x}{21s}+\tfrac{y}{20s}=1 \ \iff \ 20x+21y-420s$.

Using the classical formula of the distance of a point to a straight line, we have: $d(I,(H))=\tfrac{|20t+21t-420s|}{\sqrt{20^2+21^2}}$ which must be equal to $t$, yielding finally the following constraint:

$$\tag{2}\tfrac{420s-41 t}{29}=t.$$

The linear system (1)+(2) gives instantly : $s=2$ and $t=12$.

Thus $OA=21s=42$ and $OB=20s=40$ and the area is $\frak{A}$$=\tfrac12 \times 42 \times 40=840.$

Remark: the problem is based on the pythagorean triangle $(20,21,29)$.

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