Let $ABC$ be a non-equilateral triangle and let $R$ be the radius of its circumcircle. The incircle of $ABC$ has $I$ as its centre and is tangent to side $CA$ in point $D$ and to side $CB$ in point $E$. Let $A_1$ be the point on line $EI$ such that $A_1I=R$, with $I$ being between $A_1$ and $E$. Let $B_1$ be the point on line $DI$ such that $B_1I=R$, with $I$ being between $B_1$ and $D$. Let $P$ be the intersection of lines $AA_1$ and $BB_1$. (a) Prove that $P$ belongs to the circumcircle of $ABC$. (b) Let us now also suppose that $AB=1$ and $P$ coincides with $C$. Determine the possible values of the perimeter of $ABC$.
Problem
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Tags: geometry
Seicchi28
11.05.2022 13:31
Part (a) Similar setup with this. To prove $P \in \odot(ABC)$, it suffices to show that $\measuredangle(AA_1, BB_1) = \measuredangle(AC, BC)$, or equivalently, $\measuredangle(AA_1, AC) = \measuredangle(BB_1, BC)$. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(200), C = dir(-20), O = circumcenter(A, B, C); pair I = incenter(A, B, C), N = dir(90), L = -N; pair ai = foot(I, B, C); pair I1 = 2*ai - I; pair AA = intersectionpoints(circumcircle(A, B, C), N--(2*I-N))[1]; pair lol = N + I; pair I2 = extension(I, I1, N, lol); pair M = intersectionpoints(circumcircle(A, B, C), N--(I2+100*(N-I2)))[0]; pair D = intersectionpoints(circumcircle(A, B, C), M--(M+(100*(I-M))))[1]; pair LM = (L+M)/2; draw(A--B--C--cycle, red+linewidth(1)); draw(A--L^^L--O^^L--M^^O--LM, magenta); draw(circumcircle(A, B, C), heavymagenta); draw(circumcircle(A, I, M), gray+dashed); draw(incircle(A, B, C), gray); draw(I2--I^^D--M, gray+dashed); draw(I2--M^^O--I, green); draw(I--I2^^O--N, blue); string[] names = {"$A$", "$B$", "$C$","$A_2$", "$I$","$N$","$O$", "$A_1$","$M$","$L$"}; pair[] points = {A, B, C,D, I,N,O,I2,M,L}; pair[] ll = {A, B, C,D, I, N, O,I2,M,L}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy]Let $L, N$ be the midpoint of minor and major arc $BC$ of $\odot(ABC)$, respectively. Since $IA_1=ON=R$ and they are parallel, so $IONA_1$ is a parallelogram, i.e. $OI \parallel A_1N$. Let $A_1N$ intersects $\odot(ABC)$ for the second time at $M$, so $LM \perp MN$. Therefore, $OI \perp MN$. Let $A_2$ be the second intersection of $MI$ and $\odot(ABC)$. This means $AA_2 \perp OI$. Observe that: \[ \measuredangle(AIA_1) = \measuredangle(ALN) = \measuredangle(AMN) = \measuredangle(AMA_1), \]so $A, A_1, M, I$ are concyclic. Therefore: (Here, we denote $\widehat{PQ}$ as the inscribed angle subtended by directed arc $PQ$) \begin{align*} \measuredangle(AA_1, AC) &= \measuredangle(AA_1, AI)+\measuredangle(IAC) \\ &= \measuredangle(NMI) + \measuredangle(LAC) \\ &= \widehat{NA_2} + \widehat{LC} \\ &= \widehat{NB} + \widehat{BA_2} + \widehat{BL} = 90^{\circ} + \widehat{BA_2}. \end{align*}Similarly, by letting $B_2$ as a point on $\odot(ABC)$ such that $OI \perp BB_2$, we obtain: $\measuredangle(BB_1, BC) = 90^{\circ} + \widehat{AB_2}$. Since $AA_2 \parallel BB_2$, we obtain $\widehat{AB_2} = \widehat{BA_2}$. Therefore, the two angles are equal as desired, and the conclusion follows. $\square$ Part (b) If $P \equiv C$, therefore $A_1 \in AC$. Therefore, from the right triangle $\triangle CEA_1$, since $I \in EA_1$ and $CI$ bisects $\angle ECA_1$, we obtain: \[ \cos(C) = \frac{EC}{CA_1} = \frac{EI}{IA_1} = \frac{r}{R}, (\star) \]where $r$ is the inradius of $\triangle ABC$. Using $\cos(C) = (a^2+b^2-c^2)/2ab$, $r=\Delta/s$, $R=abc/4\Delta$, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$, and $s=(a+b+c)/2$ to expand $(\star)$ gives us the following quadratic equation on $c$: \[ c^2 - c \left( \frac{2ab}{a+b} \right) - (a-b)^2 = 0. \]Let $x = a+b, y=ab$. We have the estimate $x^2 \ge 4y$, and $x = a+b > c = 1$ from triangle inequality. Since $c=1$ is a root, therefore by $abc$ formula: \[ \frac{y}{x} \pm \sqrt{\frac{y^2}{x^2} + x^2 - 4y} = 1. \]Obviously, the negative sign couldn't be possible because then the left-hand side would be a non-positive number. Therefore, \[ \frac{y}{x} + \sqrt{\frac{y^2}{x^2} + x^2 - 4y} = 1 \rightarrow \frac{x^2-1}{2-\frac{1}{x}} = 2y \le \frac{x^2}{2}. \]The last inequality gives us $2 \ge x$. If $x=2$, then $4y=x^2 \rightarrow a=b=1 \rightarrow \triangle ABC$ is equilateral, impossible. Therefore, $x < 2$. So, we have: \[ 1 < x < 2 \rightarrow 2 < \text{Perimeter}(ABC) = a+b+c=x+1 < 3. \]So, the possible values of perimeter of $\triangle ABC$ lies on the interval $(2, 3)$.