Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 2 down vote accepted

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 $$ \frac{\lvert\overline{AB}\rvert+\lvert\overline{BC}\rvert-\lvert\overline{AC}\rvert}2=\frac{\sqrt{17}-1+\sqrt{17}+1-6}2=\sqrt{17}-3\tag{6} $$

share|improve this answer

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.

share|improve this answer
1  
This one might be illustrative as well: mathschallenge.net/full/… –  M.B. Oct 31 '12 at 17:08

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 $r$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.