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

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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$

$AG = GC$

$AF = BF$

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

4 Answers 4

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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) :

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

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    $\begingroup$ Thanks for help $\endgroup$ May 5, 2023 at 12:33
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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.

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  • $\begingroup$ Thank you, but it would be an exercise to be solved by plane geometry. $\endgroup$ May 2, 2023 at 20:51
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$\triangle ABC$: 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)

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  • $\begingroup$ I edited my answer. I showed your formula is correct. Also your answer is perfect. +1. $\endgroup$
    – sirous
    May 5, 2023 at 6:50
  • $\begingroup$ @sirous Inspired by this answer of peta arantes, I have provided one which IMHO is a little more direct. $\endgroup$
    – Jean Marie
    May 5, 2023 at 7:38
  • $\begingroup$ Haven't you a little typo right at the beginning : $\angle A = 2A$ instead of $\angle A = \tfrac12 A$ ? $\endgroup$
    – Jean Marie
    May 5, 2023 at 7:39
  • $\begingroup$ @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 $\angle CAN=\angle A = \angle NAB$ $\endgroup$ May 5, 2023 at 12:43
  • $\begingroup$ Yes. I understand now your conventions. $\endgroup$
    – Jean Marie
    May 5, 2023 at 13:37
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enter image description here

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.

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

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