BrMO 2 2008

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Source: Question 2

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SimonM

#1 h Jan 31, 2008, 8:50 PM • 1 Y

Y by Adventure10

Let triangle ABC have incentre $I$ and circumcentre $O$ . Suppose that $AIO = 90^\circ$ and $CIO = 45^\circ$ . Find the ratio $AB: BC: CA$ .

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#2 h Jan 31, 2008, 11:58 PM • 2 Y

Y by Adventure10, Mango247

It is well known, and also posted in this forum, that condition $\angle AIO=90$ implies $2a=b+c$ .

Since $\angle CIA=135$ we conclude that $B=90$ .

Now, by sine law, equation $2a=b+c$ can be transformed into its equivalent:

$2\sin(A)-1=\cos(A)$ and after squaring this becomes equivalent to: $\sin(A)[5\sin(A)-4]=0$ and this one implies that: $\sin(A)=\frac{4}{5}$ and consequently $\sin{C}=\frac{3}{5}$ and therefore:

$a: b: c=4: 5: 3$

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#3 h Feb 18, 2008, 9:54 AM • 2 Y

Y by Adventure10, Mango247

delegat wrote:

It is well known, and also posted in this forum, that condition $\angle AIO = 90$ implies $2a = b + c$ .

Where is it?

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#4 h Feb 18, 2008, 10:00 AM • 2 Y

Y by Adventure10, Mango247

See here: http://www.mathlinks.ro/viewtopic.php?t=156016, and for more equivalences, see Virgil Nicula's posts from: http://www.mathlinks.ro/viewtopic.php?t=164835.

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campos

#5 h Feb 28, 2008, 2:43 AM • 2 Y

Y by Adventure10, Mango247

the problem states that $\angle CIO=45$ or $\angle CIA=45$ ?

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#6 h Feb 3, 2015, 10:14 PM • 2 Y

Y by Adventure10, Mango247

[url=http://www.bmoc.maths.org/home/bmo2-2008.pdf]2007/8 British Mathematical Olympiad Round 2[/url] Problem 2 wrote:

Let triangle $ABC$ have incentre $I$ and circumcentre $O$ . Suppose that $\angle AIO=90^o$ and $\angle CIO=45^o$ . Find the ratio $AB:BC:CA$ .

Let $\angle A=\alpha$ , $\angle B=\beta$ and $\angle C=\gamma$ . Considering the angles of $\triangle{ACI}$ , paired with the fact that $I$ lies on all three interior angle bisectors, we have: $135^o+\frac{\alpha}{2}+\frac{\gamma}{2}=180^o\Leftrightarrow \beta=90^o.$ Also, let $AI=a$ , $CI=c$ , $OI=\ell$ and $AO=CO=BO=R$ ; hence, applying the cosine rule to $\triangle{AOI}$ , $\triangle{OCI}$ and $\triangle{ACI}$ yields (resp.): $a^2+\ell^2=R^2\quad (*)$ $\ell^2+c^2-\ell c \sqrt{2} = R^2\quad (**)$ $a^2+c^2+ac\sqrt{2}=4R^2\quad (***)$ From (*) and (**) we get $R^2-\ell^2$ $=a^2$ $=c^2-\ell c\sqrt{2}$ $\Leftrightarrow \ell = \tfrac{c^2-a^2}{c\sqrt{2}}$ , and plugging this back into (*) and simplifying gives $R^2=\tfrac{a^4+c^4}{2c^2}$ $=\tfrac{a^4}{2c^2}+\tfrac{c^2}{2}$ $\Leftrightarrow 4R^2$ $=\tfrac{2a^4}{c^2}+2c^2$ . Therefore, substituting this for the RHS of (***) we have: $a^2+c^2+ac\sqrt{2}=\frac{2a^4}{c^2}+2c^2$ $\Leftrightarrow \left(\frac{a}{c}\right)^2+\left(\frac{a}{c}\right)\sqrt{2} = 2\left(\frac{a}{c}\right)^4+1$ $\Leftrightarrow 2u^4-u^2-u\sqrt{2}+1=0,$ where $u=\tfrac{a}{c}$ . Now, to rid ourselves of the irrational coefficient, we let $u=v\sqrt{2}$ to obtain: $8v^4-2v^2-2v+1=0,$ at which point the rational root theorem yields an easy factorization, namely, $(2v-1)^2(2v^2+2v+1)=0.$ So the only real (and positive) root is $v=\tfrac{1}{2}$ , since $\Delta_{2,2,1}$ $=-4<0$ . It follows that $u=\tfrac{\sqrt{2}}{2}$ is the root we are after. Then, by the extended sine rule on $\triangle AIB$ , followed by $\beta=90^o$ (and similarly for $\triangle BIC$ ), we can easily compute that: $a=2\sqrt{2}R \sin (\tfrac{\gamma}{2})\ \text{and}\ c=2\sqrt{2}R\sin (\tfrac{\alpha}{2}),$ which, upon squaring and using (***) and the fact that $\sin^2 x=\tfrac{1-\cos 2x}{2}$ , gives: $AB:AC=\frac{4R^2-c^2}{4R^2}=\frac{a^2+ac\sqrt{2}}{a^2+c^2+ac\sqrt{2}}=\frac{u^2+u\sqrt{2}}{u^2+u\sqrt{2}+1};$ and similarly, $BC:AC=\frac{4R^2-a^2}{4R^2}=\frac{u\sqrt{2}+1}{u^2+u\sqrt{2}+1}.$ Therefore, $AB:BC:CA=u^2+u\sqrt{2}:u\sqrt{2}+1:u^2+u\sqrt{2}+1,$ which, upon substituting in the value of $u$ we found earlier, becomes: $AB:BC:CA=\frac{3}{2}:2:\frac{5}{2}=\boxed{3:4:5}.$ Q.E.D.

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