It's problem 3 from the first Romanian TST in 2002. I have a computational proof and I won't post it because it's really boring: we use Menelaus to find the ratio in which side BC is divided by AA₁, then we write Ceva's condition for concurrence with the ratios we have just calculated and we get something like (tanA-tanB)(tanB-tanC)(tanC-tanA)=0, so 2 of them must be equal, so the triangle is isosceles. I remember slaving on this one for 45 minutes or so (during th contest).

Valiowk wrote: Let M and N be the midpoints of the respective sides AB and AC of an acute-angled triangle ABC. Let P be the foot of the perpendicular from N onto BC and let A₁ be the midpoint of MP. Points B₁ and C₁ are obtained similarly. Well, this looks a bit strange since the definition of $A_1$ is not symmetric in $B$ and $C$. However, we can easily see that if we take the foot $Q$ of the perpendicular from $M$ onto $BC$, and define $A_1$ as the midpoint of $NQ$, then we get the same point $A_1$. In fact, the quadrilateral $MNQP$ is a rectangle. Valiowk wrote: If $AA_1$, $BB_1$ and $CC_1$ are concurrent, show that triangle $ABC$ is isosceles. Well, I have reduced this to $(b^2+3c^2-a^2)(c^2+3a^2-b^2)(a^2+3b^2-c^2)$ = $(c^2+3b^2-a^2)(a^2+3c^2-b^2)(b^2+3a^2-c^2)$ ==> triangle is isosceles. But this is algebra, again. Any nice proof? Darij

I think I got here as well during that contest. We replace expressions of the type a²+b²-c² with 2ab*cosC, we divide both sides by 8 and by abc, and we get \pi{cyl}(b*cosA+c)=\pi{cyl}(c*cosA+b)=>\pi{cyl}(b*cosA+c)/(a*cosB+c)=1. b*cosA+a*cosB=c and b*cosA/a*cosB=tanB/tanA=>(b*cosA+c)/(a*cosB+c)=(2tanB+tanA)/(2tanA+tanB)=>\pi{cyl}(2tanB+tanA)/(2tanA+tanB)=1, and from here it's easy to derive \pi{cyl}(tanA-tanB)=0. P.S. {cyl} comes from cyclic.

See the attached drawing, the triangles $ MNA_1$ and $ BCA_2$ are centrally similar, centered at $ A$. $ A_2$ is the center of the rectangle erected outwardly on $ BC$, of dimensions $ a, h_a$. If $ A' \equiv BC \cap AA_2$, we have then: $ \frac {BA'}{CA'} = \frac {S_{[ABA_2]}}{S_{[ACA_2]}}=\frac{AB \cdot BA_2 \cdot sin (\widehat {ABC}+\widehat {CBA_2})}{AC \cdot CA_2 \cdot sin (\widehat {ACB}+\widehat {BCA_2})}=\frac{c \cdot sin (\angle B + \angle \alpha)}{b \cdot sin (\angle C + \angle \alpha )}$ ( 1 ), where $ a, b, c$ and $ h_a, h_b, h_c$ are the sides and altitudes of the triangle, $ S_{[XYZ]}$ is the area of the $ \triangle XYZ$, $ \alpha$ being the half-angle of the diagonals of the rectangle constructed as explained and other two similar relations. If $ AA', BB', CC'$ are concurrent ($ B', C'$ are obtained similarly), then, by Ceva we get as the product of fractions (1) the unit and from this we get the same results as Darij. Best regards, sunken rock

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Computational solution, although I'll post it since it seems sort of instructive on ``how to bash'' Not that there's anything terribly new here... Employ barycentric coordinates. Let $S_A = \frac{b^2+c^2-a^2}{2}$, et cetera. By taking the midpoint of the foot of the perpendicular $(0: S_C : S_B)$ with $C = (0 : 0 : a^2)$ we find that $P=(0 : S_C : S_B + a^2)$. With $M = (a^2 : a^2 : 0)$ we get $A_1 = \left( a^2 : S_C+a^2 : S_B+a^2 \right)$. Hence $AA_1$ divides $BC$ in a $S_C+a^2$ to $S_B+a^2$ ratio. Now by Ceva's Theorem, the condition is equivalent to \begin{align*} 0 &= (S_B+a^2)(S_C+b^2)(S_A+c^2)-(S_B+c^2)(S_C+a^2)(S_A+b^2) \\ &= \sum_{\text{cyc}} a^2(b^2-c^2)S_a + \sum_{\text{cyc}} (c^2-b^2)S_{BC} \\ &= \sum_{\text{cyc}} (b^2-c^2)(a^2S_A-S_{BC}) \\ &= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(2a^2(b^2+c^2-a^2)-(b^2+a^2-c^2)(c^2+a^2-b^2)) \\ &= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(2a^2b^2+2b^2c^2+2c^2a^2+b^4+c^4-3a^4) \\ &= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(b^4+c^4-3a^4) \\ &= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(a^4+b^4+c^4-4a^4) \\ &= - \sum_{\text{cyc}} a^4(b^2-c^2) \\ &= (a^2-b^2)(b^2-c^2)(c^2-a^2) \end{align*} which holds if and only if $ABC$ is isosceles.

Valiowk wrote: Let $M$ and $N$ be the midpoints of the respective sides $AB$ and $AC$ of an acute-angled triangle $ABC$. Let $P$ be the foot of the perpendicular from $N$ onto $BC$ and let $A_1$ be the midpoint of $MP$. Points $B_1$ and $C_1$ are obtained similarly. If $AA_1$, $BB_1$ and $CC_1$ are concurrent, show that the triangle $ABC$ is isosceles. Mircea Becheanu Is there a full proof

Can somebody help me

Valiowk wrote: ...Points $B_1$ and $C_1$ are obtained similarly... Mircea Becheanu This is not a clear definition for $B_1, C_1.$

Solved from LiOG 3.8 Let the foot from $M$ to $BC$ be $T$, so $A_1$ is the center of rectangle $MNPT$. Let $AA_1$ meet $MN$ at $R$ and $BC$ at $S$. Let $MR=d$. Due to symmetry, since $AA_1$ passes through the center of the rectangle, we have $SP=d$ as well. Furthermore, by homothety at $A$, $BS=2d$, so $BP=3d$. However, we also have $$BP=BT+TP=\frac{c}{2}\cos\beta+\frac{a}{2},$$so therefore, $$3d=\frac{c}{2}\cos\beta+\frac{a}{2}.$$Thus, $$d=\frac{c\cos\beta+a}{6},$$so $$\frac{BS}{CS}=\frac{\frac{c\cos\beta+a}{3}}{a-\frac{c\cos\beta+a}{3}}=\frac{a+c\cos\beta}{2a-c\cos\beta}=\frac{2a^2+2ac\cos\beta}{4a^2-2ac\cos\beta}.$$However, by Law of Cosines, this is equal to $$\frac{3a^2+c^2-b^2}{3a^2+b^2-c^2}.$$Thus, by Ceva's theorem, it suffices to show that $$\prod_{cyc}(3a^2+b^2-c^2)=\prod_{cyc}(3a^2-b^2+c^2)$$implies that $\triangle ABC$ is isosceles. Let $p=a^2,q=b^2,c=r^2$ so that this is $$\prod_{cyc}(3p+q-r)=\prod_{cyc}(3p-q+r).$$Let $x=q-r,y=r-p,z=p-q$ so that this is $$(3p+x)(3q+y)(3r+z)=(3p-x)(3q-y)(3r-z)$$Since terms with an even number of factors in $\{x,y,z\}$ cancel, this is just $$9(pqz+pry+qrx)+xyz=0.$$Substituting $x,y,z$ back in, this is $$9(\sum_{cyc}pq(p-q))+(p-q)(q-r)(r-p)=\sum_{cyc}8p^2q+p^2r-9pq^2,$$and evaluating the cyclic sum and setting it to 0, this is simply just $$8(p^2q+q^2r+r^2p)-8(p^2r+q^2p+r^2q),$$so the condition is just $$p^2q+q^2r+r^2p-(p^2r+q^2p+r^2q)=0.$$This factors as $$(p-q)(q-r)(r-p)=0,$$so two of $p,q,r$ are equal so $\triangle ABC$ is isosceles and we are finally done as we step back from this crazy bash.