Find the measure of the radius of the circle inscribed in the triangle $IMN$

In the figure, calculate the measure of the radius of the circle inscribed in the triangle $$IMN$$, where $$a$$ and $$b$$ are arrows and $$c$$ is the radius of the semicircumference . Remarks: $$O$$ is $$I$$ are respectively circumcenter and incenter of triangle $$ABC$$.

(S:$$\frac{\sqrt{abc}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$$)

I try

$$\triangle OMN: MN^2 = c^2+c^2 \implies MN =c\sqrt2$$

$$\angle BAM \cong \angle BCM \cong \angle\cong \angle MNA$$

$$\angle BAN \cong \angle BAC \cong \angle ANO \cong \angle CMN$$

$$\triangle MIN \sim \triangle AIC(A.A.)\implies \frac{c\sqrt2}{c}=\sqrt2 = k$$

$$AG = GC$$

$$AF = BF$$

• Add in your condictions that points where perpendicularity are show, are midpoints!!! May 2, 2023 at 16:10
• A property that can help $c=a+b+\sqrt{a.b}$ May 2, 2023 at 21:12
• @petaarantes: why is that? May 3, 2023 at 6:46
• @Dominique It is a property of metric relations on the circle May 3, 2023 at 14:00
• @Dominique : In fact, it is a consequence of this relationship : $(c-a)^2+(c-b)^2=c^2$ (Pythagoras in triangle $OGM$). May 5, 2023 at 6:38

Let us enumerate different "facts" :

1. Pythagoras theorem in triangle $$OGM$$ gives :

$$(c-a)^2+(c-b)^2=c^2 \tag{1}$$

1. In triangle $$MIN$$, we have the (classical) relationship

$$xp=S\tag{2}$$

between its inradius $$x$$, its half-perimeter $$p$$ and its area $$S$$.

1. $$AMI$$ and $$CNI$$ are right isosceles triangles.

Explanation for triangle $$AMI$$ : $$\angle AMI=90°$$ because it subtends a diameter and $$\angle MAI=45°=90°/2$$ because it subtends arc MBN which is subtended by the center angle $$\angle MON$$. (similar proof for triangle $$CNI$$).

As a consequence

$$AM=IM=\sqrt{2ac} \ \text{and} \ CN=IN=\sqrt{2bc}\tag{3}$$

Let us prove the second relationship in (3) (it is the same reasoning for the first relationship).

Pythagoras theorem in triangle $$CGN$$ gives, using (1)

$$CN^2=b^2+(c-a)^2=b^2+c^2-(c-b)^2=2bc$$

As a consequence of (3), the half perimeter of triangle $$IMN$$ is :

$$p=\frac12 \left(\sqrt{2ac} + \sqrt{2bc}+ c \sqrt{2} \right)=\sqrt{c}\frac{\sqrt{2}}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right)\tag{4}$$

Besides, it is rather easy to prove that angle $$\angle MIN = 135°$$ giving, for the area of triangle $$MIN$$ :

$$S=\frac12 IM.IN \sin 135°=\frac12 \sqrt{2ac}\sqrt{2bc} \frac{\sqrt{2}}{2}=c\sqrt{ab}\frac{\sqrt{2}}{2}\tag{5}$$

Now, we can use (2) with expressions coming from (3) and (5) :

$$x=\frac{S}{p}=\frac{c\sqrt{ab}\frac{\sqrt{2}}{2}}{\sqrt{c}\frac{\sqrt{2}}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right)}=\frac{\sqrt{abc}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$$

• Thanks for help May 5, 2023 at 12:33

HINT.-Let $$B=(x_0,y_0),A=(-c,0),C=(c,0)$$. Midpoint of $$AC=(0,0)$$, of $$AB=(\dfrac{x_0-c}{2},\dfrac{y_0}{2})$$, of $$BC=(\dfrac{x_0+c}{2},\dfrac{y_0}{2})$$.

(1) Incenter: $$I=\left(\dfrac{-c(BC)+c(AB)+2cx_0}{AB+BC+2c},\dfrac{2cy_0}{AB+BC+2c}\right)$$ by known formula easy to deduce.

(2) $$N$$ point: it is given by the system $$y=\dfrac{y_0}{x_0+c}x\\x^2+y^2=4c^2$$ and similarly $$M$$ point is solution of $$y=\dfrac{y_0}{x_0-c}x\\x^2+y^2=4c^2$$ (3) Calculate sides $$MN,MI$$ and $$NI$$ then using again the formula above we have the incenter, say $$J$$, of the triangle $$\triangle{MNI}$$.

(4) We have the radius $$x$$ which is equal, for instance, to the distance from $$J$$ point to the line $$MN$$. So we get an explicit function of $$(a,b,c,x_0,y_0)$$. We can eliminate $$x_0$$ and $$y_0$$ from the two equalities $$b(2c-b)=\left(\frac{BC}{2}\right)^2\\a(2c-a)=\left(\frac{AB}{2}\right)^2$$ which are very known by beginners.

• Thank you, but it would be an exercise to be solved by plane geometry. May 2, 2023 at 20:51

$$\triangle ABC$$: Let $$\angle A = 2A$$ e $$\angle C = 2C \implies \angle A + \angle C = 45^o$$

$$\angle IMN = \angle CMN = \angle A = \angle IAC$$

similarly $$\angle INM = \angle C = \angle ICA \therefore \triangle MIN \sim \triangle AIC$$

$$\angle IAM = \angle NAM = \angle C + \angle A = 45^{\circ}$$

$$IM = AM = \frac{AI}{\sqrt2} \implies k =\sqrt2$$

$$MN = c\sqrt2$$

$$AM = 2c sin \angle C$$

$$NI=2c sin \angle A$$

$$sin (2A) = \frac{c-a}{c}\\ sin (2C) = \frac{c-b}c\\ \therefore sin (A) = \sqrt{\frac a{2c}}\\ AM = 2c \cdot sin (C) = \sqrt{2ac} \implies AI = 2 \sqrt{ac}\\ x \sqrt{2} = \frac{S}{p} = \frac{2S}{2p} = \frac{AI \cdot CI \cdot sin (135^{\circ})}{2p}\\ \therefore 2x = \frac{4c\sqrt{ab}}{2p} \implies x = \frac{2c\sqrt{ab}}{2c + 2\sqrt{bc} + 2\sqrt{ac}} = \boxed{\frac{\sqrt{abc}}{\sqrt a +\sqrt b + \sqrt c }}$$

(S:FelipeMartin)

• @sirous Inspired by this answer of peta arantes, I have provided one which IMHO is a little more direct. May 5, 2023 at 7:38
• Haven't you a little typo right at the beginning : $\angle A = 2A$ instead of $\angle A = \tfrac12 A$ ? May 5, 2023 at 7:39
• @JeanMarie If $\angle A = \frac{A}{2}$ and $C= \frac{C}{2}$ we will have $A+C = 45^o$ which is not true. I called $\angle CAN=\angle A = \angle NAB$ May 5, 2023 at 12:43
• Yes. I understand now your conventions. May 5, 2023 at 13:37

Hint: Calulation from my figure shows that $$x\approx \frac r3$$, where r is the radius of inscribed circle of triangle ABC. In my figure , $$a=7.31$$, $$b=30.65$$ and $$c=60$$ and calculations by your formula gives $$x=7.3\approx 2\times 14.6$$ which is what the figure show.So the formula is correct.

One way can be finding the measure of sides of triangle MNI, say m, n and i. then use this formula:

$$x=\frac Sp$$

wher s is the area and p is half perimer of triangle MNI.

1- to find IM, first find angle $$ACB=\gamma$$ by sine rule. Then find CM. we have:

$$CI=\frac r{sin \frac {\gamma}2}$$

$$MI=CM-CI$$

Similarly find NI. You found $$MN=c\sqrt 2$$. These are enough for calculation of x.

Note:

$$\angle BAC=\alpha$$

$$\angle ABC=\beta$$

$$\angle ACB=\gamma$$

You could segments label GH and MF some thing other than a and b to orevent misunderstanding.

• My figure on geogebra shows that the result is correct...See the figure I posted May 2, 2023 at 19:46
• @petaarantes, I gave the measures of the sides of ABC. You can calculate yourself. May 3, 2023 at 6:40
• Which program did you use to generate this image? May 3, 2023 at 16:46
• @冥王Hades. Solidworlss, it is for mechanical engineering. May 5, 2023 at 6:22