In a triangle $ABC$, it is known that $ AC = m$, $AB = 3m$, $BC = Rm$.

Find for which values of $R $ the triangle is:

(i) An acute angle triangle.

(ii) A right angle triangle.

(iii) An obtuse angle triangle.

I have tried several approaches including the law of cosine and the Pythagorean theorem, but it didn't help.


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Hint: first to find when the triangle will be right triangle.($E,F$ is trivial)

Then check $A,B,C$ ,which kind of triangle it will form. the rest is easy.

Edit: following is full answer:

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$B$ will on the circle green with radius $3m$, $AC=m<AB-AC=2m<BC<4m=AB+AC$

So if $\triangle ABC$ is right triangle, then $AC$ will be the leg . to be the right triangle, the only possible position for B is on $D$ or $E$, $\implies R=\sqrt{10} $ or $ R=\sqrt{8}$ for (ii)

if $B$ on between $DE$ , note the $BH$ is altitude, so $\angle BAC$ and $\angle BCA$ will be acute angle, $AC$ is the shortest side, so $\angle ABC$ must be acute angle . so in this case , $\triangle ABC$ must be (i).

If$B$ is on $EG$ or $DF$, note $H_1$ or $H_2$ will be outside of $AC$, so $ $\angle BCA$ or $\angle BAC$ must be obtuse angle , that is case (iii)

the last step is to find $R$'s range.

let$AH=x \implies BC^2= BH^2+HC^2=AB^2-AH^2+(|AH-AC|)^2=Ab^2+AC^2-2AH*AC=10m^2-2mx \implies R^2=10-\dfrac{2x}{m}$

so $R$ is mono decreasing fonction of $x$

$ \sqrt{8} <R <\sqrt{10}$ for case(i)

$R >\sqrt{10}$ or $R<\sqrt{8}$ for case(iii)

  • $\begingroup$ I'm afraid I cant feel that triviality. $\endgroup$ – Bak1139 Jan 26 '14 at 13:20
  • 1
    $\begingroup$ @Bak1139 $3m>m$, so $m$ must be the leg. so there is only two possibility for the right triangle, either $3m$ is leg or $Rm$ is leg. those are the case $E,F$.The $A,B,C$ here is not your triangle $ABC$ $\endgroup$ – chenbai Jan 26 '14 at 13:27
  • $\begingroup$ awesome work, thank you! $\endgroup$ – Bak1139 Jan 28 '14 at 16:40

You can use the law of cosines 9 times, since we are unsure which angle will be the right or obtuse in those cases. For the sake of simplicity, lets let $AB=c, AC=b, BC=a$. Choose angle $A$ to be our first case; by the law of Cosines, $$a^2=b^2+c^2-2bc\cos{(A)}$$ The three cases are: if $A\lt{90}$ degrees, if $A=90$ degrees, and if $A\gt{90}$ degrees.

If $A\lt{90}$, then $\cos{(A)}$ is positive which means $$(Rm)^2=m^2+(3m)^2-2m(3m)\cos{(A)}$$ $$R^2m^2=10m^2-6m^2\cos{(A)}$$ Thus $R=\sqrt{10-6\cos{A}}.$

If $A=90$ degrees, then $\cos{(A)}=0$ which means $$R^2m^2=m^2+9m^2$$ Thus $R=\sqrt{10}$.

Finally if $A\gt{90}$ degrees, then $\cos{(A)}$ is negative which $$R^2m^2=m^2+9m^2+2m(3m)\cos{(A)}$$ $$R^2m^2=10m^2+6m^2\cos{(A)}$$ Thus $R=\sqrt{10+6\cos{(A)}}$

Repeat this argument with $b^2=a^2+c^2-2ac\cos{(B)}$ and $c^2=a^2+b^2-2ab\cos{(C)}$.

EDIT: with the other two cases, you will end up with quadratics in $R$, since the law of cosines with produce an $R^2$ term and an $R$ term. For example, $$m^2=9m^2+R^2m^2-18R\cos{(B)}$$ Now you must solve $$R^2-18\cos{(B)}R+8=0$$ When $B=90$ degrees it just becomes solving $R^2+8=0$ which produces complex solutions, not what we are after...

  • $\begingroup$ this makes sense,although Im a little worried it may take a little to much time ,on a test, for instance. $\endgroup$ – Bak1139 Jan 26 '14 at 13:39
  • $\begingroup$ This is the general case: you don't know what sides are adjacent to which angles so you are forced to attempt them all (unless of course some one has a much simpler way to solve). My guess is that you would be given the explicit angle which could be acute, right, or obtuse... $\endgroup$ – Eleven-Eleven Jan 26 '14 at 13:48
  • $\begingroup$ No, actually this is a test question $\endgroup$ – Bak1139 Jan 26 '14 at 13:49
  • $\begingroup$ Oh, well perhaps there's a simpler way.... I'm not sure of one myself.... $\endgroup$ – Eleven-Eleven Jan 26 '14 at 14:09

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