# Triangle Calculation Question

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. 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. $B$ 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) • I'm afraid I cant feel that triviality. – Bak1139 Jan 26 '14 at 13:20 • @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$ABC$– chenbai Jan 26 '14 at 13:27 • awesome work, thank you! – 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...

• this makes sense,although Im a little worried it may take a little to much time ,on a test, for instance. – Bak1139 Jan 26 '14 at 13:39
• 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... – Eleven-Eleven Jan 26 '14 at 13:48
• No, actually this is a test question – Bak1139 Jan 26 '14 at 13:49
• Oh, well perhaps there's a simpler way.... I'm not sure of one myself.... – Eleven-Eleven Jan 26 '14 at 14:09