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A circle is inscribed in a right angled triangle ABC where AC is the hypotenuse. The circle touches AC at point P. Length of AP = 2unit and CP = 4 units.

What is the radius of the circle?

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In the following diagram

$\hspace{3.5cm}$enter image description here

we know that $\lvert\overline{AP}\rvert=2$ and $\lvert\overline{CP}\rvert=4$. Therefore, we also have $\lvert\overline{AQ}\rvert=2$ and $\lvert\overline{CR}\rvert=4$.

We know by the Pythagorean Theorem that $$ \lvert\overline{AB}\rvert^2+\lvert\overline{BC}\rvert^2=\left(\lvert\overline{AP}\rvert+\lvert\overline{CP}\rvert\right)^2=36\tag{1} $$ and since $\lvert\overline{QB}\rvert=\lvert\overline{RB}\rvert$, $$ \lvert\overline{BC}\rvert-\lvert\overline{AB}\rvert=\lvert\overline{CR}\rvert-\lvert\overline{AQ}\rvert=2\tag{2} $$ Solving $(1)$ and $(2)$ for $\lvert\overline{AB}\rvert$ and $\lvert\overline{BC}\rvert$ yields $$ \lvert\overline{AB}\rvert=\sqrt{17}-1\qquad\text{and}\qquad\lvert\overline{BC}\rvert=\sqrt{17}+1\tag{3} $$ Therefore, the radius of the incircle equals $$ \lvert\overline{RB}\rvert=\lvert\overline{BC}\rvert-\lvert\overline{CR}\rvert=\sqrt{17}-3\tag{4} $$


Alternatively, the radius of the incircle is the area of the triangle divided by the semiperimeter. Thus, the inradius is $$ \frac{\frac12(\sqrt{17}-1)(\sqrt{17}+1)}{\frac12(\sqrt{17}-1+\sqrt{17}+1+6)} =\frac8{\sqrt{17}+3}=\sqrt{17}-3\tag{5} $$


We can use the formula for the inradius of a right triangle cited by M.B. to verify that the inradius is $$ \frac{\lvert\overline{AB}\rvert+\lvert\overline{BC}\rvert-\lvert\overline{AC}\rvert}2=\frac{\sqrt{17}-1+\sqrt{17}+1-6}2=\sqrt{17}-3\tag{6} $$

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Let $Q$ and $R$ be the points where the circle touches $AB$ and $BC$, respectively. Then $|AP| = |AQ| = 2$ and $|CP| = |CR| = 4$. Also note that $|AB| = |AQ| + r$ where $r$ is the radius of the circle. Similarly we get $|BC| = |CR| + r$. Hence $$6^2 = |AC|^2 = |AB|^2 + |BC|^2 = (2+r)^2 + (4+r)^2 = (4 + 4r + r^2)+(16 + 8r + r^2)$$ $$= 20 + 12r + 2r^2$$

This you can solve.

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Let the radius be $r$, and let $O$ be the centre of the incircle.

Recall that the incentre is where the angle bisectors meet. Let $2\alpha$ be the angle at $A$, and $2\gamma$ the angle at $C$. Then $\angle OAP=\alpha$ and $\angle OCP=\gamma$.

Note that $2\alpha+2\gamma=\pi/2$, so $\alpha+\gamma=\pi/4$.

Note also that $\tan\alpha=\dfrac{r}{2}$ and $\tan\gamma=\dfrac{r}{4}$.

By the addition law for tangent, we have $$1=\tan(\alpha+\gamma)=\frac{\tan\alpha+\tan\gamma}{1-\tan\alpha\tan\gamma}.\tag{$1$}$$ But $\tan\alpha=2\tan\gamma$. So from Equation $1$ we get $$1-2\tan^2\gamma=3\tan\gamma.$$ This is a quadratic equation in $\tan\gamma$. Solve, and use $\tan\gamma=\dfrac{r}{4}$ to find $r$.

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