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# can anybody solve this?

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The right triangle with sides 3, 4, 5 has an incircle with radius 1.  There is another triangle with sides 3, 4, x that also has an incircle with radius 1.  What is x?

Jul 13, 2020

#1
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The right triangle with sides 3, 4, 5 has an incircle with radius 1.

There is another triangle with sides 3, 4, x that also has an incircle with radius 1.

What is x?

The inradius  r of the incircle in a triangle with sides of length  a, b , c is given by

$$r = \dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$, where $$s=\dfrac{a+b+c}{2}$$

$$\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s},} & \mathbf{s=\dfrac{a+b+c}{2}} \\\\ && \boxed{r=1,\ a=3,\ b=4,\ c=x } \\\\ 1 &=& \dfrac{\sqrt{s(s-3)(s-4)(s-x)}}{s}, & s=\dfrac{3+4+x}{2} \\\\ s &=& \sqrt{s(s-3)(s-4)(s-x)}\quad \text{square both sides} , & s=\dfrac{7+x}{2} \\\\ s^2 &=& s(s-3)(s-4)(s-x) \\\\ s &=& (s-3)(s-4)(s-x) \\\\ \dfrac{7+x}{2} &=& \left(\dfrac{7+x}{2}-3 \right) \left(\dfrac{7+x}{2}-4 \right) \left(\dfrac{7+x}{2}-x \right) \\\\ \dfrac{7+x}{2} &=& \dfrac{(x+1)(x-1)(7-x)}{8} \\\\ 4 ( 7+x ) &=& (x+1)(x-1)(7-x) \\\\ 28+4x &=& (x^2-1)(7-x) \\\\ 28+4x &=& 7x^2-x^3-7+x \\ \mathbf{x^3-7x^2+3x+35} &=& \mathbf{0} \\ && \boxed{x=5, \text{ see the right triangle with sides }3,\ 4,\ \mathbf{5}} \\\\ (x-5)(\underbrace{x^2-2x-7}_{=0}) &=& 0 \\ \hline \mathbf{x^2-2x-7} &=& \mathbf{0} \\\\ x &=& \dfrac{2\pm \sqrt{4-4*(-7)}}{2} \\\\ x &=& \dfrac{2\pm \sqrt{32}}{2} \\\\ x &=& \dfrac{2\pm \sqrt{16*2}}{2} \\\\ x &=& \dfrac{2\pm 4\sqrt{2}}{2} \\\\ \mathbf{ x } &=& \mathbf{ 1\pm 2\sqrt{2} } \\ \hline \\ \mathbf{ x} &=& \mathbf{1+2\sqrt{2}} \\ x &=& 1-2\sqrt{2} \quad x < 0 !,\ \text{no solution!} \\ \hline \end{array}$$

x is $$\mathbf{1+2\sqrt{2}} = 3.82842712475$$ Jul 13, 2020
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Thank you, Guest ! heureka  Jul 13, 2020