Inequality with areas

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Inequality with areas

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Source: Moldova 2007 IMO-BMO TST I problem 1

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iura

#1 h Mar 5, 2007, 4:16 PM • 2 Y

Y by Adventure10 and 1 other user

Let $ABC$ be a triangle and $M,N,P$ be the midpoints of sides $BC, CA, AB$ . The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$ . Show that the area of triangle $ABC$ is at most the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$ .

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#2 h Mar 5, 2007, 5:20 PM • 2 Y

Y by Adventure10 and 1 other user

We have $\frac{[A_{1}BC]}{[ABC]}=\frac{A_{1}M}{AM},...$ etc, therefore we only need prove $\frac{A_{1}M}{AM}+\frac{B_{1}N}{BN}+\frac{C_{1}M}{CM}\geq 1.$

But this statement is easy, i think so

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edriv

#3 h Mar 5, 2007, 6:03 PM • 2 Y

Y by Adventure10, Mango247

I'm going to use the fact that the symmetric of the ortocentre wrt. a side lies on the circumcircle.

Let's call $H_{A},H_{B},H_{C}$ the intersections of the altitudes with the circumference; $D,E,F$ the intersections of the altitudes with the sides.

Since the baricentre lies between the ortocentre and the circumcentre, and the last ones are isogonal conjugate, we have that $C_{1}$ is further than $H_{C}$ from the line $AB$ . Then $(ABC_{1}) \ge (ABH_{C}) = (ABH)$ .
Summing up, we get:
$(ABC_{1})+(BCA_{1})+(CAB_{1}) \ge (ABH)+(BCH)+(CAH) = (ABC)$
QED

@ n.t.tuan: could you explaing why your inequality is easy?

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#4 h Mar 5, 2007, 9:43 PM • 2 Y

Y by Adventure10, Mango247

very nice solution

edriv wrote:

@ n.t.tuan: could you explaing why your inequality is easy?

If $ABC$ has an obtuse angle, it is obvious.
Otherwise if you call $A_{0}$ the midpoint of the smallest arc BC and $A'$ the midpoint of the greatest arc BC, then you have $\frac{A_{1}M}{AM}\ge \frac{A_{0}M}{A'M}= tan^{2}\frac{\alpha}{2}$ .
Since tan² is convex we have $tan^{2}\frac{\alpha}{2}+tan^{2}\frac{\beta}{2}+tan^{2}\frac{\gamma}{2}\ge 3 tan^{2}30 = 1$ , qed.

Benjamin

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#5 h Mar 8, 2007, 10:23 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

iura wrote:

Let $ABC$ be a triangle and $M,N,P$ be th midpoints of sides $BC, CA, AB$ . The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$ . Show that the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$ is at least equally to the area of the triangle $ABC$ .

Proof. Suppose that the triangle $ABC$ hasn't an obtuse angle. Denote the circumcircle $w=C(O,R)$ of the given triangle, the area $S=[ABC]$ and the points $A'\in w\cap(OM$ , $B'\in w\cap (ON$ , $C'\in w\cap (OP$ . Observe that $MA_{1}\ge MA'$ and $MA'=OA'-OM=R(1-\cos A)$ a.s.o. Therefore, $\sum\frac{[BA_{1}C]}{[ABC]}=$ $\sum\frac{MA_{1}}{MA}=$ $\sum\frac{MA^{2}_{1}}{MA\cdot MA_{1}}\ge$ $\sum\frac{A'M^{2}}{MB\cdot MC}=$ $\sum \frac{4R^{2}(1-\cos A)^{2}}{a^{2}}=$ $\sum\frac{4R^{2}(1-\cos A)^{2}}{4R^{2}(1-\cos^{2}A)}=$ $\sum\frac{1-\cos A}{1+\cos A}=$ $\sum f(A)\ge 3\cdot f\left(\frac{A+B+C}{3}\right)=$ $3\cdot f\left(\frac{\pi}{3}\right)=$ $3\cdot\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=1$ , where $f: \left(0,\frac{\pi}{2}\right]\rightarrow \mathrm R$ , $f(x)=\frac{1-\cos x}{1+\cos x}=\tan^{2}\frac{x}{2}$ is a convex function (prove easily) and I applied the Jensen's inequality. In conclusion, $\sum [BA_{1}C]\ge [ABC]$ .

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#6 h Mar 9, 2007, 5:50 AM • 2 Y

Y by Adventure10 and 1 other user

We have $MA.MA_{1}=MB.MC=\frac{a^{2}}{4}$ , so $\frac{MA_{1}}{MA}=\frac{a^{2}}{4m_{a}^{2}}$ , and we need prove $\sum\frac{a^{2}}{m_{a}^{2}}\geq 4$ .

Now, by AM-GM we have $a^{2}m_{a}^{2}=\frac{4}{3}(\frac{3}{4}a^{2}).(m_{a}^{2})\leq \frac{4}{3}(\frac{\frac{3}{4}a^{2}+m_{a}^{2}}{2})^{2}=\frac{(a^{2}+b^{2}+c^{2})^{2}}{12}$ , etc.

Final, $\sum\frac{a^{2}}{m_{a}^{2}}=\sum\frac{a^{4}}{a^{2}m_{a}^{2}}\geq \frac{12(a^{4}+b^{4}+c^{4})}{(a^{2}+b^{2}+c^{2})^{2}}\geq 4$ . Done!

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#7 h Mar 12, 2007, 2:15 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Virgil Nicula wrote:

Observe that $MA_{1}\ge MA'$

Why is this?

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#8 h Mar 22, 2007, 5:23 PM • 1 Y

Y by Adventure10

barasawala wrote:

Virgil Nicula wrote:

Observe that $MA_{1}\ge MA'$

Why is this?

Given a circle $k$ and a point $P$ , how would you find the point on $k$ which lies nearest to $P$ ?

If you don't find this obvious, here is a more formal argument: $A^{\prime}\in w\cap(OM$ yields $OM+MA^{\prime}=OA^{\prime}$ . Since both points $A^{\prime}$ and $A_{1}$ lie on the circumcircle of triangle $ABC$ , and the center of this circumcircle is $O$ , we have $OA^{\prime}=OA_{1}$ . Thus, $OM+MA^{\prime}=OA_{1}$ . But the triangle inequality in $\triangle OMA_{1}$ yields $OM+MA_{1}\geq OA_{1}$ . Thus, $OM+MA_{1}\geq OM+MA^{\prime}$ , so that $MA_{1}\geq MA^{\prime}$ , qed.

darij

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