1
$\begingroup$

I'm trying to solve the following problem :

In $△ABC$, $AB = AC, BC = 48$ and inradius $r = 12$. Find the circumradius $R$.

Here is a figure that I drew : ( note : it was not given in the question so there may be some mistakes ) Figure

I don't know how to solve it , am I missing any relation between inradius , circumradius and sides of a isosceles triangle?

EDIT: Is there a simple solution without using trigonometry ?

$\endgroup$
6
  • $\begingroup$ Is that a circle through $A$, $B$, and $C$? $\endgroup$ Sep 2, 2013 at 14:28
  • $\begingroup$ @Michael, he's talking about a "circumradius", so probably. $\endgroup$
    – JMCF125
    Sep 2, 2013 at 14:31
  • 1
    $\begingroup$ Wait, $O$ is both the center of the inner and outer circles? If so, the triangle should be equilateral as well. In that case, I just figured out the solution! $\endgroup$
    – JMCF125
    Sep 2, 2013 at 14:36
  • $\begingroup$ @MichaelAlbanese Yes , at least I tried to draw a circle. $\endgroup$
    – A Googler
    Sep 2, 2013 at 15:05
  • 1
    $\begingroup$ @JMCF125 Thanks for pointing out ! But that is not the case . I'll edit the figure.. $\endgroup$
    – A Googler
    Sep 2, 2013 at 15:06

4 Answers 4

1
$\begingroup$

Denote the center of incircle by $O$ and of circumcircle by $O'$. It's easy to caclulate that $$\angle ABC=\angle ACB= 2\angle OBC=2\arctan\frac{r}{BC/2}=2\arctan\frac{1}{2}$$ Thus we can calculate the height $h$ with base $BC$ is $$h=\frac{BC}{2}\tan\angle ABC=24\tan(2\arctan\frac{1}{2})=24\times\frac{2\times\frac{1}{2}}{1-(\frac{1}{2})^2}=32$$ By symmetry, $O'$ shall lie on the height $h$. Consider the property of circumcircle that $O'A=O'B=O'C=R$ $$O'B^2=O'C^2=(\frac{BC}{2})^2+(h-R)^2=R^2$$ which gives the solution $$R=25$$

$\endgroup$
2
  • $\begingroup$ Can you give a solution without trigonometry? I didn't really understood your solution since I'm not familiar with inverse trigonometric functions etc. $\endgroup$
    – A Googler
    Sep 5, 2013 at 12:21
  • $\begingroup$ @AGoogler Don't give up. inverse trigonometric function is, as its name, gives the angle given corresponding value of trigonometric function. For example, $\tan\theta=\alpha$, then $\theta=\arctan\alpha$. Which line can't you understand? Just tell me. $\endgroup$
    – Shuchang
    Sep 5, 2013 at 15:25
1
$\begingroup$

Let $M$ be the midpoint of $BC$, let $P$ be the point where the perpendicular from $O$ meets the side $AB$, and let $|PA|=:x$. Since the two tangent segments from $B$ to the incircle have equal length it follows that $|PB|=24$; therefore $|AB|=24+x$, and $|AO|^2= 12^2+x^2$. It follows that $$(24+x)^2=24^2+\bigl(12+\sqrt{12^2+x^2}\bigr)^2\ .$$ Solving for $x$ gives $x=16$, whence $|AB|=40$, $|AO|=20$, $|AM|=32$.

Now let $|MK|=:y$. Then $\sqrt{24^2+y^2}=32-y$, which enforces $y=7$. It follows that $R=32-7=25$.

$\endgroup$
2
  • $\begingroup$ Can you please tell how is PB=24? $\endgroup$
    – A Googler
    Sep 6, 2013 at 13:05
  • $\begingroup$ @A Googler: See my edit. $\endgroup$ Sep 6, 2013 at 18:57
0
$\begingroup$

enter image description here

$AI=k$, $\triangle AIF\sim\triangle ABD$ $\Longrightarrow$ $AB=2k$ , $AD=k+12$

$AB^{2}=BD^{2}+AD^{2}$ $\Longrightarrow$ $k=20$ , $AF=s-a=16$ , $AT=20$

$\triangle AOT\sim\triangle AIF$ $\Longrightarrow$ $\dfrac{AO}{AT}=\dfrac{AI}{AF}=\dfrac{5}{4}$ $\Longrightarrow$ $AO=25$

$\endgroup$
12
  • $\begingroup$ What is s and a? How are these triangles similar? Can you simplify your solution? $\endgroup$
    – A Googler
    Sep 6, 2013 at 13:04
  • $\begingroup$ s=(a+b+c)/2, a=BC, b=CA, c=AB, O=circumcenter, I=incenter $\endgroup$
    – chloe_shi
    Sep 6, 2013 at 17:57
  • $\begingroup$ T=midpoint of AB, F=tangency point $\endgroup$
    – chloe_shi
    Sep 6, 2013 at 17:58
  • $\begingroup$ OT is perpendicular to AB and two pairs of triangles are similar $\endgroup$
    – chloe_shi
    Sep 6, 2013 at 17:59
  • $\begingroup$ by AA similarity criterion $\endgroup$
    – chloe_shi
    Sep 6, 2013 at 18:00
0
$\begingroup$

The standard formula for the inradius is r = area/s,where s is the semi-perimeter.Letting x = the triangle sides that are equal,we have s=(1/2)(x+x+48)= x+24.The area is calculated using Heron"s formula = sqrt(s(s-a)(s_b)(s_c)).In your problem,this becomes sqrt((x+24)(24)(24)(x-24)= 24sqrt(x^2 - 576). Substituting,we get: 12 = 24sqrt(x^2 - 576)/(x + 24).Dividing by 12,and cross- multiplying, x+24 = 2sqrt(x^2 - 576).Squaring,(x+24)^2 = 4(x^2 - 576),or x^2 + 48x + 576 = 4(x^2 - 576) = 4x^2 - 2304.Expanding and transposing, 3x^2 - 48x - 2880=0. Dividing by 3,x^2 - 16x - 960 =0.Solving for x, 2x = 16 + sqrt(256 + 3840),or 2x=16+sqrt(4096),and x=8 + 32 = 40.Then the area is 24sqrt(1600 - 576) = 24sqrt(1024)=24*32 = 768.A formula for the circumradius is R = abc/(4*area).In your problem,this becomes: R = (40)(40)(48)/((4)(8)(8)(12)) = (5)(%)(4)/(4) = 25. Ed Gray

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .