Let $ABC$ be a triangle, and let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$, respectively. Let $R$ be its circumradius and $r$ its inradius. Suppose that $b + c = 2a$ and $R = 3r$. The excircle relative to vertex $A$ intersects the circumcircle of $ABC$ at points $P$ and $Q$. Let $U$ be the midpoint of side $BC$, and let $I$ be the incenter of $ABC$. Prove that $U$ is the centroid of triangle $QIP$.
Problem
Source: Pan American Girls' Mathematical Olympiad 2024, P6
Tags: geo, geometry, PAGMO, complex numbers, barycentric coordinates
tobiSALT
28.11.2024 01:17
synthetic sol, solved it with Ignacio Naguil from Argentina.
Let $J$ be the $A$-excenter.
Let $D$ be the midpoint of arc $BC$ not containing $A$.
By Ptolemy's theorem in $ABDC$, $(AB+AC)BD=AD \times BC$. Since $b+c=2a$, $AD=2BD$, but it is known that $DI=DB=DJ$, so $AI=ID=DJ$.
Let $T$ and $G$ be the tangency points of $AB$ and $BC$ with the incircle, respectively. Since $AT=s-a=a/2=BU$ and $AI=BD$, and $\angle BAD = \angle DAC = \angle DBU$, triangles $ATI$ and $BUD$ are congruent. In particular $DU=r$.
Since $GI=r=DU$, we have that $ID$ intersects $GU$ at its midpoint ($M$). Moreover, if we call $K$ the tangency point of the $A$-excircle with $BC$, we know that $GU=UK$, so $MK=3MU$ and by similarity in $\triangle MUD$ and $\triangle MKJ$, $JK=3r$.
Let $X$ be the symmetric of $I$ with respect to $U$. If we prove that $PUQX$ is a parallelogram we win. In particular, if we prove that the midpoints of $UX$ and $PQ$ are the same, we win.
First, notice that the circumcircle of $ABC$ and the $A$-excircle have the same radius, so $OPJQ$ is a rhombus and the midpoints of $OJ$ and $PQ$ coincide.
By the reflection, $TX=r$. Then $JX=2r$. Since $OU=OD-DU=2r$, we have $JX=OU$ and also $OU \parallel JX$, so $OUJX$ is a parallelogram and the midpoints of $OJ$ and $UX$ coincide.
Therefore, the midpoints of $OJ$, $UX$ and $PQ$ coincide. Particularly, the midpoints of $UX$ and $PQ$ coincide, as we wanted.
Attachments:
hukilau17
28.11.2024 01:26
"If the only tool you have is a hammer, you'll treat everything as if it's a nail." We proceed by complex bashing. Identify the $A$-excircle of $\triangle ABC$ with the unit circle, and let it touch sides $\overline{BC},\overline{CA},\overline{AB}$ at $D,E,F$ respectively. Then (letting $j$ denote the coordinate of $I$) we have $$|d|=|e|=|f|=1$$$$a=\frac{2ef}{e+f}$$$$b=\frac{2df}{d+f}$$$$c=\frac{2de}{d+e}$$$$u = \frac{b+c}2 = \frac{d(de+df+2ef)}{(d+e)(d+f)}$$$$j = \frac{4def}{(d+e)(d+f)}$$Now let $z$ be the coordinate of either $P$ or $Q$. Since $z$ lies on the $A$-excircle, we have $|z|=1$. Since $A,B,C,Z$ are concyclic, we have $$\frac{(a-b)(c-z)}{(a-c)(b-z)} \in \mathbb{R} \implies \frac{f(c-z)}{e(b-z)} \in \mathbb{R}$$$$\frac{f(d+f)(2de-dz-ez)}{e(d+e)(2df-dz-fz)} = \frac{e(d+f)(2z-d-e)}{f(d+e)(2z-d-f)}$$$$f^2(2z-d-f)(dz+ez-2de) = e^2(2z-d-e)(dz+fz-2df)$$$$2f^2z^2(d+e) - f^2z(d^2+df+ef+5de) + 2def^2(d+f) = 2e^2z^2(d+f) - e^2z(d^2+de+ef+5df) + 2de^2f(d+e)$$$$2z^2(e-f)(de+df+ef) - z(e-f)(d^2e+d^2f+de^2+6def+df^2+e^2f+ef^2) + 2def(e-f)(d+e+f) = 0$$$$2z^2(de+df+ef) - z(d^2e+d^2f+de^2+6def+df^2+e^2f+ef^2) + 2def(d+e+f) = 0$$Thus by Vieta, we have $$p+q = \frac{d^2e+d^2f+de^2+6def+df^2+e^2f+ef^2}{2(de+df+ef)}$$Now since $AB + AC = 2BC$, we have $$\frac{a-b}f + \frac{c-a}e = \frac{2(c-b)}d$$$$\frac{2e}{e+f}-\frac{2d}{d+f}+\frac{2d}{d+e}-\frac{2f}{e+f} = \frac{4e}{d+e}-\frac{4f}{d+f}$$$$e(d+e)(d+f) - d(d+e)(e+f) + d(d+f)(e+f) - f(d+e)(d+f) = 2e(d+f)(e+f) - 2f(d+e)(e+f)$$$$(d+e)(d+f)(e-f) - d(e+f)(e-f) = 2d(e+f)(e-f)$$$$(d+e)(d+f) = 3d(e+f)$$$$\boxed{d^2-2de-2df+ef = 0}$$Also, $R = 3r$. Now we have (since $\triangle DEF$ is obtuse) $$R = \frac2{|d+e+f|^2-1} = \frac{2def}{(d+e)(d+f)(e+f)}$$and $$r = \frac{j-2d+d^2\overline{\jmath}}{2d} = \frac{(d-e)(d-f)}{(d+e)(d+f)}$$Then $$\frac{2def}{(d+e)(d+f)(e+f)} = \frac{3(d-e)(d-f)}{(d+e)(d+f)}$$$$2def = 3(d-e)(d-f)(e+f)$$$$3d^2e+3d^2f-3de^2-8def-3df^2+3e^2f+3ef^2 = 0$$Plugging in $d^2 = 2de+2df-ef$ simplifies this to $$3de^2+4def+3df^2 = 0 \implies \boxed{3e^2+4ef+3f^2 = 0}$$Now we write $$(d+e)(d+f) = 3d(e+f)$$so that $$j = \frac{4ef}{3(e+f)}$$$$u = \frac{de+df+2ef}{3(e+f)}$$Applying $d^2 = 2de+2df-ef$ to the numerator of $p+q$ as well gives $$p+q = \frac{3de^2+10def+3df^2}{2(de+df+ef)}$$and applying $3e^2+4ef+3f^2 = 0$ then gives $$p+q = \frac{3def}{de+df+ef}$$Meanwhile, $$3u-j = \frac{3de+3df+2ef}{3(e+f)}$$Then $$3u-p-q-j = \frac{(3de+3df+2ef)(de+df+ef) - 9def(e+f)}{3(e+f)(de+df+ef)} = \frac{3d^2e^2+6d^2ef+3d^2f^2-4de^2f-4def^2+2e^2f^2}{3(e+f)(de+df+ef)}$$Applying $3e^2+4ef+3f^2=0$ to the numerator gives $$3u-p-q-j = \frac{2ef(d^2-2de-2df+ef)}{3(e+f)(de+df+ef)}$$which is zero since $d^2-2de-2df+ef=0$. Thus $3u-p-q-j=0$, so $$u = \frac{p+q+j}3$$as desired. $\blacksquare$
MathLuis
28.11.2024 02:15
Let $AI \cap (ABC)=N$, let incircle, and A-excircle touch $BC$ at $D,Z$ respectively, also let $H$ in $I_AZ$ such that $NH \parallel BC$ (where $I_A$ is the A-excenter). let $s$ be the semi-perimeters then from areas $s \cdot r=[ABC]=(s-a)r_a$ where $r_a$ is radius of the A-excircle, this means in general as $r=1.5a$ we get that $R=3r=r_a$ which means that the A-excircle and $(ABC)$ are congruent which gives that if $O$ is the center of $(ABC)$ then $OPI_AQ$ is a rhombus, also note that if you let $ZH=x$ then from I-E lemma as we have $IN=NI_A$ ee also get that $r+x=3r-x$ meaning that $x=r$ which means that $BICH$ is a parallelogram which means $H$ is orthocenter of $\triangle BI_AC$, but more importantly we have that since $ZI_A=ON$ that $OZI_AN$ is a paralelogram and as $UN=ZH=r$ we have that $OHI_AU$ is a parallelogram which means that $QP,OI_A,UH$ share the same midpoint $M$ , because of the parallelogram $BICH$ we have $I,U,H$ colinear and in fact $IU=UH$ but also from the other common midpoint you get the desired ratio $2:1$ on lenght considering $U$ in $IM$, therefore $U$ is barycenter of triangle $\triangle PIQ$ as desired, thus we are done .
hectorleo123
28.11.2024 03:46
Very easy by bash
Let $\omega$ be the circumcircle of triangle $ABC$ and $\Omega$ the $A$-excircle
Let $O$ be the circumcenter of $ABC$, $I_a$ the $A$-excenter and $T$ a point on the ray $IU$ such that $\frac{IU}{UT}=2$
$$S=\frac{abc}{4R}=sr\Rightarrow \frac{abc}{4R}=\frac{3aR}{6}\Rightarrow R^2=\frac{bc}{2}$$$$S=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}\Rightarrow R^2=\frac{a^2b^2c^2}{16s(s-a)(s-b)(s-c)}$$$$\Rightarrow 8s(s-a)(s-b)(s-c)=a^2bc\Rightarrow 8(\frac{3a}{2})(\frac{a}{2})(s-b)(s-c)=a^2bc\Rightarrow 6(s-b)(s-c)=bc$$$$\Rightarrow 6(3c-b)(3b-c)=16bc\Rightarrow 9b^2+9c^2=22bc\ldots (\alpha)$$We will use barycentric coordinates with $A=(1,0,0), B=(0,1,0)$ and $C=(0,0,1)$
$$\Rightarrow U=(0:1:1), I=(a:b:c), I_a=(-a:b:c) \text{ and } O=(a^2S_a:b^2S_b:c^2S_c)$$$T=(x_t:y_t:z_t)\Rightarrow \frac{2x_t+a}{3}=0, \frac{2y_t+b}{3}=\frac{a+b+c}{2}$ and $\frac{2z_t+c}{3}=\frac{a+b+c}{2} \Rightarrow T=(-2a:3a+b+3c:3a+3b+c)$
$\omega:$ $-a^2yz-b^2zx-c^2xy=0$ and $\Omega: -a^2yz-b^2zx-c^2xy+(x+y+z)(s^2x+(s-c)^2y+(s-b)^2z)=0$
$$\Rightarrow PQ: s^2x+(s-c)^2y+(s-b)^2z=0$$Claim 1: $T\in PQ$
Proof: $T\in PQ\Leftrightarrow -2as^2+(3a+b+3c)(s-c)^2+(3a+3b+c)(s-b)^2=0$
$$\Leftrightarrow s^2(4a+4b+4c)-(3a+b+3c)c(2s-c)-(3a+3b+c)b(2s-b)=0$$$$\Leftrightarrow (a+b+c)((a+b+c)^2-c(a+b)-b(a+c))-2(a+b)(a+c)c-2(a+b)(a+c)b=0$$$$\Leftrightarrow (a+b+c)(a^2+b^2+c^2+ab+ac)=2(a+b)(a+c)(b+c)$$$$\Leftrightarrow 3a(a^2+b^2+c^2+ab+ac)=4a(a+b)(a+c)$$$$\Leftrightarrow 3b^2+3c^2+\frac{3(b+c)^2}{2}=a^2+4ab+4ac+4bc$$$$\Leftrightarrow12b^2+12c^2+6b^2+12bc+6c^2=(b+c)^2+16bc+8(b+c)^2$$$$\Leftrightarrow 9b^2+9c^2=22bc \text{ by }(\alpha) \text{ we are done }_\blacksquare$$
$$T=(-2a:3a+b+3c:3a+3b+c)=(-b-c:\frac{5b+9c}{2}:\frac{9b+5c}{2})=(-2b-2c:5b+9c:9b+5c)$$$$I_a=(-a:b:c)=(-b-c:2b:2c)$$$$32a^2S_a=(b+c)^2(4b^2+4c^2-(b^2+2bc+c^2))=(b+c)^2(3b^2+3c^2-2bc)$$$$32b^2S_b=4b^2(5c^2+2bc-3b^2)=Y, 32c^2S_c=4c^2(5b^2+2bc-3c^2)=Z$$$$\Rightarrow O=((b+c)^2(3b^2+3c^2-2bc):Y:Z)$$Claim 2: $T\in OI_a$
Proof: $T, O, I_a$ are collinear
$$\Leftrightarrow
\begin{vmatrix}
-2b-2c & 5b+9c & 9b+5c\\
-b-c & 2b & 2c\\
(b+c)^2(3b^2+3c^2-2bc) & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow
\begin{vmatrix}
-2 & 5b+9c & 9b+5c\\
-1 & 2b & 2c\\
(b+c)(3b^2+3c^2-2bc) & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow
\begin{vmatrix}
-4 & 10b+18c & 18b+10c\\
-1 & 2b & 2c\\
(b+c)(3b^2+3c^2-2bc) & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow
\begin{vmatrix}
1 & 18c & 18b\\
-1 & 2b & 2c\\
(b+c)(3b^2+3c^2-2bc) & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow
\begin{vmatrix}
3 & 18c & 18b\\
-3 & 2b & 2c\\
(b+c)(9b^2+9c^2-6bc) & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow
\begin{vmatrix}
1 & 6c & 6b\\
-3 & 2b & 2c\\
16(b+c)bc & Y & Z
\end{vmatrix}
=0$$$$\Leftrightarrow 192(b^4c+b^3c^2-b^2c^3-bc^4)+Y(2c+18b)-Z(18c+2b)=0$$$$\Leftrightarrow 96(b^4c+b^3c^2-b^2c^3-bc^4)=Z(9c+b)-Y(9b+c)$$$$\Leftrightarrow 96(b^4c+b^3c^2-b^2c^3-bc^4)=4c^2(5b^2+2bc-3c^2)(9c+b)-4b^2(5c^2+2bc-3b^2)(9b+c)$$$$\Leftrightarrow 24(b^4c+b^3c^2-b^2c^3-bc^4)=c^2(5b^2+2bc-3c^2)(9c+b)-b^2(5c^2+2bc-3b^2)(9b+c)$$$$\Leftrightarrow 24(b^4c+b^3c^2-b^2c^3-bc^4)=(45b^2c^3+5b^3c^2+18bc^4+2b^2c^3-27c^5-3bc^4)-(45c^2b^3+5c^3b^2+18cb^4+2c^2b^3-27b^5-3cb^4)$$$$\Leftrightarrow 24(b-c)(bc(b^2+bc+c^2)+b^2c^2)=(b-c)(-45b^2c^2+5b^2c^2-18bc(b^2+bc+c^2)-2b^2c^2+27(b^4+b^3c+b^2c^2+bc^3+c^4)+3bc(b^2+bc+c^2))$$$$\Leftrightarrow 24bc(b+c)^2=(-45b^2c^2+5b^2c^2-18bc(b^2+bc+c^2)-2b^2c^2+27(b^4+b^3c+b^2c^2+bc^3+c^4)+3bc(b^2+bc+c^2))$$$$\Leftrightarrow 24bc(b+c)^2=(-45b^2c^2+5b^2c^2-18bc(b^2+bc+c^2)-2b^2c^2+27((b^2+c^2)^2+b^3c+bc^3-b^2c^2)+3bc(b^2+bc+c^2))$$$$\Leftrightarrow 24bc(b+c)^2=(-45b^2c^2+5b^2c^2-18bc(b^2+bc+c^2)-2b^2c^2+27((\frac{22bc}{9})^2+b^3c+bc^3-b^2c^2)+3bc(b^2+bc+c^2))$$$$\Leftrightarrow 24(b+c)^2=(-45bc+5bc-18(b^2+bc+c^2)-2bc+27((\frac{484bc}{81})+b^2+c^2-bc)+3(b^2+bc+c^2))$$$$\Leftrightarrow 72(b+c)^2=(-126bc-54(b^2+bc+c^2)+81((\frac{484bc}{81})+b^2+c^2-bc)+9(b^2+bc+c^2))$$$$\Leftrightarrow 72(b+c)^2=(-126bc-54(b^2+bc+c^2)+(484bc+81b^2+81c^2-81bc)+9(b^2+bc+c^2))$$$$\Leftrightarrow 72(b+c)^2=36b^2+36c^2+232bc$$$$\Leftrightarrow 36b^2+36c^2=88bc \text{ by }(\alpha) \text{ we are done }_\blacksquare$$Note that since $P$ and $Q$ are the intersection points of $\omega$ and $\Omega$, then $OI_a$ is the perpendicular bisector of $PQ$.
Then $T$ is the midpoint of $PQ$, and by the definition of $T$ we have that $U$ is the centroid of triangle $IPQ_\blacksquare$.
ohhh
28.11.2024 20:39
consider tobiSALTI's figure, and $h =$ lenght of $A-$altitude. Easily we get $h = R = r_a$, then $OPQI_a$ is a rhombus $\implies p + q = o + j\implies i + p + q = o + i + j$. Considering $(ABC)$ the unit circle, $i + p + q = 2d$, so we only want $2d = 3u$, i.e., $2R = 3OU$, $\iff r = UD \iff MI = MD$, but: • $h = r_a \implies MA = MJ$ • $\frac{MI}{IA} = \frac{MD}{MJ}$ (just trig, it's always true) $\implies MI = MD$, as desired. Nice problem
Thelink_20
29.11.2024 20:07
Synthetic FTW Define $r_a, p, O, I_a$ as usual. Let $D$ be the contact of the incircle and $AC$ and $N, M$ midpoints of $OI_a$, $\overarc{BC}$. $rp=r_a(p-a)\implies r_a=3r=R\implies\boxed{POQI_a \ rhombus}\implies N$ midpoint of $PQ. \ (*)$. But of course: $\Delta ADI\sim\Delta CUM\implies \frac{DI}{AD}=\frac{UM}{UC}\implies\frac{r}{p-a}=\frac{UM}{\frac{a}{2}}\implies UM=r=\frac{R}{3}\implies\boxed{OU:UM=2:1}$ And it's well known that $M$ bissects $II_a$, so that means $U$ is the centroid of $\Delta OII_a\implies I-U-N$ and $IU:UN=2:1\implies U$ is the centroid of $\Delta QIP$, because of $(*)$ .$_{\blacksquare}$
Thelink_20
29.11.2024 23:08
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