Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Problem
Source: 2016 IMO Shortlist G4
Tags: geometry, incenter, reflection, IMO Shortlist, Inversion, Cyclic Quadrilaterals
19.07.2017 19:38
Let $I'$ be the reflection of $I$ through $AC$. Let $E’ \equiv AI \cap (BDI’)$ and $X \equiv I’C \cap AI$. Notice that $BXI’D$ is cyclic. Furthermore $\angle BE’C= 2 \angle BE’X=2 \angle BI’X$. Also $E’$ lies on the perpendicular bisector of $BC$, which implies that $E’$ must be the circumcenter of $BI’C$. Thus $BE’=E’I$, and $BE’DI$ is cyclic, which implies $E’D \perp AB$. And we conclude $E \equiv E’$.
19.07.2017 19:45
This is a pretty direct application of complex numbers. We'll take the incircle as the unit circle, tangent to $BC$, $AC$, $AB$ at $1$, $z$, $1/z$. Line $BI$ coincides with a diameter $UV$ of the unit circle for which $u+v=0$ and $uv = -1/z$, and so we have \[ d = \frac{uv(z+z) - zz(u+v)}{uv-zz} = \frac{-2}{-1/z-z^2} = \frac{2z}{z^3+1}. \]To compute $e$, we have \begin{align*} \frac{e-a}{0-a} &= \frac{d-a}{z-a} \implies \frac{e}{-a} = \frac{d-z}{z-a} \\ e &= \frac{z-d}{z-a} \cdot a = \frac{z - \frac{2z}{z^3+1}}{z - \frac{2z}{z^2+1}} \cdot \frac{2z}{z^2+1} \\ &= 2z^2 \cdot \frac{1 - \frac{2}{z^3+1}}{z(z^2-1)} \\ &= \frac{(z^3-1)(2z^2)}{z(z^2-1)(z^3+1)} = \boxed{\frac{2z(z^2+z+1)}{(z^3+1)(z+1)}}. \end{align*}Now the reflection of $I$ in $\overline{AC}$ is $2z$, so we have \begin{align*} \frac{2z-d}{\frac{2}{z+1}-d} \div \frac{2z-e}{\frac{2}{z+1}-e} &= \frac{2z \cdot \left[ 1 - \frac{1}{z^3+1} \right]} {\frac{2}{z+1} \left[ 1 - \frac{z}{z^2-z+1} \right]} \div \frac{2z \left[ 1-\frac{z^2+z+1}{(z+1)(z^3+1)} \right]} {\frac{2}{z+1} \left[ 1 - \frac{z(z^2+z+1)}{z^3+1} \right]} \\ &= \frac{1 - \frac{1}{z^3+1}} {1 - \frac{z}{z^2-z+1}} \div \frac{1-\frac{z^2+z+1}{(z+1)(z^3+1)}} {1 - \frac{z(z^2+z+1)}{z^3+1}} \\ &= \frac{z^3}{(z+1)(z^2-2z+1)} \div \frac{(z+1)(z^3+1) - (z^2+z+1)} {(z+1)\left[ (z^3+1) - z(z^2+z+1) \right]} \\ &= \frac{z^3}{(z-1)^2} \div \frac{z^2(z^2+z-1)}{-z^2-z+1} = \frac{-z}{(z-1)^2} \end{align*}which is evidently a real number.
19.07.2017 20:07
Let $XYZ$ be the triangle formed by the midpoints of minor arcs in $\odot(ABC)$. First notice that $\overline{CI'Y}$ are collinear and that lines $CY, AX, BZ$ are concurrent a point $W$; we claim that $W\in \odot(BDI')$. Just compute $$\angle BDI'=2\angle BDC=2(\pi-\angle BCY)=\pi-\angle BWC$$Finally, we claim that $E\in\odot(BWD)$. Compute $$\angle BDE=\angle AIB-\angle DEI=\bigg(\frac{\pi+\angle C}{2}\bigg)-\bigg(\frac{\pi+\angle A}{2}\bigg)=\frac{3\angle C-\pi}{2}$$and $$\angle BWE=\frac{\angle BWC}{2}=\frac{\pi-2\angle BDC}{2}=\frac{3\angle C-\pi}{2}$$as desired.
19.07.2017 21:00
If there is an isosceles triangle, then consider coordinate bashing. (I will spare the reader many of the computations, though.) Let $B=(-1,0)$, $C=(1,0)$, $I=(0,r)$, with $0<r<1$ and $r\neq\frac{1}{\sqrt{3}}$. Also, let $P$ be the reflection of $I$ over $AC$ and $X$ be the midpoint of $IP$. By double-angle, $A=\left(0,\frac{2r}{1-r^2}\right)$. The equations of $AC$ and $BI$ are \begin{align*} 2rx+(1-r^2)y &= 2r,\\ y &= r(x+1), \end{align*}respectively. Intersecting these, $$D = \left( \frac{1+r^2}{3-r^2},\frac{4r}{3-r^2} \right).$$$IP$ has slope $\frac{1-r^2}{2r}$, so its equation is $$(1-r^2)x-2ry=-2r^2.$$Intersecting this with $AC$, $X=\left(\frac{2r^2}{r^2+1},\frac{2r}{r^2+1}\right)$, so $$P = \left( \frac{4r^2}{r^2+1},\frac{3r-r^3}{r^2+1} \right).$$$DE$ is parallel to $IP$, so its equation is $$(1-r^2)x-2ry=\frac{1-8r^2-r^4}{3-r^2},$$giving $$E = \left( 0,\frac{r^4+8r^2-1}{2r(3-r^2)} \right).$$ Let $\ell_1,\ell_2,\ell_3$ be the perpendicular bisectors of $BD,BE,BP$, respectively. $BD$ has slope $r$ and midpoint $\left(\frac{r^2-1}{3-r^2},\frac{2r}{3-r^2}\right)$, so the equation of $\ell_1$ is $$x+ry = \frac{3r^2-1}{3-r^2}.$$$BE$ has midpoint $\left(-\frac{1}{2},\frac{r^4+8r^2-1}{4r(3-r^2)}\right)$ and slope $\frac{r^4+8r^2-1}{2r(3-r^2)}$, so $\ell_2$ has equation $$2r(3-r^2)x+(r^4+8r^2-1)y = \frac{r^8+12r^6+86r^4-52r^2+1}{4r(3-r^2)}.$$Finally, $BP$ has slope $\frac{3r-r^3}{5r^2+1}$ and midpoint $\left(\frac{3r^2-1}{2(r^2+1)},\frac{3r-r^3}{2(r^2+1)}\right)$, so $\ell_3$ has equation $$(5r^2+1)x+(3r-r^3)y = \frac{r^6+9r^4+7r^2-1}{2(r^2+1)} = \frac{r^4+8r^2-1}{2}.$$ We wish to show that $\ell_1,\ell_2,\ell_3$ concur, so $BEDP$ is cyclic. Since $\ell_1$ is not horizontal, it suffices to show that $\ell_1\cap\ell_2$ and $\ell_1\cap\ell_3$ have the same $y$-coordinate. The equation of $\ell_1$ is equivalent to $x=\frac{3r^2-1}{3-r^2}-ry$. Thus, the $y$-coordinate of $\ell_1\cap\ell_2$ satisfies $$(3r^4+2r^2-1)y = \frac{r^8+12r^6+86r^4-52r^2+1}{4r(3-r^2)}-2r(3r^2-1),$$and the $y$-coordinate of $\ell_1\cap\ell_3$ satisfies $$(2r-6r^3)y = \frac{r^4+8r^2-1}{2} - \frac{(5r^2+1)(3r^2-1)}{3-r^2}.$$These $y$-coordinates are the same if and only if $$(3r^4+2r^2-1)\left(\frac{r^4+8r^2-1}{2}-\frac{(5r^2+1)(3r^2-1)}{3-r^2}\right) = (2r-6r^3)\left(\frac{r^8+12r^6+86r^4-52r^2+1}{4r(3-r^2)}-2r(3r^2-1)\right),$$which is equivalent to $$(r^2+1)\left(\frac{(r^4+8r^2-1)(3-r^2)}{2}-(5r^2+1)(3r^2-1)\right) = -\frac{r^8+12r^6+86r^4-52r^2+1}{2}+4r^2(3r^2-1)(3-r^2)$$after multiplying through by $\frac{3-r^2}{3r^2-1}$. Now both sides are equal to $\frac{-r^8-36r^6-6r^4+28r^2-1}{2}$ as desired.
19.07.2017 21:31
Let $I'$ be the reflection of $I$ across $AC$ and $II' \cap AC \equiv K$ and $I'E \cap AC \equiv L$ Let $\angle I'IB = x$ , $\angle IBI' =y$ and $\angle LIK =z$ See that , $\angle DEI' = \angle LI'K = \angle LIK$ (so it suffices to show that $sin y = sin z$) Now by angle bisector thm and RHS similarity, $\frac{DL}{LK}= \frac{EL}{I'L} = \frac{AE}{AI'} = \frac{AE}{AI} = \frac{AD}{AIcos(\frac{A}{2})}$ Now, $\frac{II'}{IB} = \frac{2rsin(\frac{B}{2})}{r} = 2sin(\frac{B}{2}) = = \frac{sin(IBI')}{sin(II'B)} = \frac{sin(y)}{sin(y+x)}$ Now,by sin law in $\triangle DIK$, $\frac{sin(z)}{sin(z+x)} = \frac{LK.ID}{KI.DL} = \frac{AI.IDcos(\frac{A}{2})}{AD.IK} = \frac{cos(\frac{A}{2})}{cos(\frac{C}{2})} = 2sin(\frac{B}{2}) $ So, $\frac{sin(z)}{sin(z+x)} = \frac{sin(y)}{sin(y+x)} => cot(z) = cot(y)$ and we are done!
20.07.2017 01:40
20.07.2017 02:07
We use barycentric coordinates to solve this problem. For simplicity, let $(a, b, c)$ denote the point $(a : b : c)$. From now on, $a = BC, b = AC, c = AB$. Furthermore, we have $b = c$ and $a \neq b$. First, we quote the formula for the reflection of a point $P = (u, v, w)$ with respect to the triangle $ABC$. Let $P^{(a)}, P^{(b)}, P^{(c)}$ be the reflections of $P$ across $BC, AC, AB$ respectively. Then, $P^{(a)} = (-a^{2}u, (a^{2} + b^{2} - c^{2})u + a^{2}v, (c^{2} + a^{2} - b^{2})u + a^{2}w)$ $P^{(b)} = ((a^{2} + b^{2} - c^{2})v + b^{2}u, -b^{2}v, (b^{2} + c^{2} - a^{2})v + b^{2}w)$ $P^{(c)} = ((c^{2} + a^{2} - b^{2})w + c^{2}u, (b^{2} + c^{2} - a^{2})w + c^{2}v, -c^{2}w)$ Now, we are ready to solve the problem. Let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$. Then, $I = (a, b, c) = (a, b, b)$ and $D = (\frac{a}{a + b}, 0, \frac{b}{a + b})$, as by the angle bisector theorem $\frac{DC}{DA} = \frac{a}{b}$ and thus $[BDC] : [ADC] : [ABD] = a : 0 : b$. Next, we find $I'$. This is easy as we can just substitute the coordinates of $I$ in the formula above to get $I' = (ab(a + b), -b^{3}, 3b^{3} - a^{2}b)$. Now, we need to find $E$. $E$ lies on line $AI$, and thus $E$ can be written in the form $(x, y, y)$ where $x + 2y = 1$. (The equation of a line passing through $A$ is $y = kz$ and since $I$ is on $AI$, substituting $y = z = b$ gives $k = 1$) $\vec{ED} = (x - \frac{a}{a + b}, y, y - \frac{b}{a + b})$ and $\vec{AC} = (1, 0, -1)$. By EFFT, $ED \perp AC$ implies $0 = a^{2}(-y) + b^{2}(\frac{a - b}{a + b} + y - x) + b^{2}(y)$ $\Rightarrow b^{2}x = y(-a^{2} + 2b^{2}) + \frac{b^{2}(a - b)}{a + b}$. Substituting $x = 1 - 2y$ and simplifying, we get $b^{2}(1 - 2y) = y(-a^{2} + 2b^{2}) + \frac{b^{2}(a - b)}{a + b}$ $\Rightarrow(a^{2} - 4b^{2})y = \frac{-2b^{3}}{a + b}$ $\Rightarrow y = \frac{2b^{3}}{(a + b)(4b^{2} - a^{2})}$ Substituting into $x = 1 - 2y$, we have $x = \frac{a(-a^{2} - ab + 4b^{2})}{(a + b)(4b^{2} - a^{2})}$ Note that $4b^{2} - a^{2} = (2b - a)(2b + a) \neq 0$ by triangle inequality. Thus, $E = \left(\frac{a(-a^{2} - ab + 4b^{2})}{(a + b)(4b^{2} - a^{2})}, \frac{2b^{3}}{(a + b)(4b^{2} - a^{2})}, \frac{2b^{3}}{(a + b)(4b^{2} - a^{2})}\right) = (a(-a^{2} - ab + 4b^{2}), 2b^{3}, 2b^{3})$. It remains to prove that $B, D, E, I'$ are concyclic. Let $\omega = -a^{2}yz - b^{2}zx - c^{2}xy + (ux + vy + wz)(x + y + z) = 0$ be the circle passing through $B, D, I'$, where $u, v, w \in \mathbb{R}$. We just have to show that $E$ lies on it. Substituting the point $B = (0, 1, 0)$ into the equation for $\omega$ gives $v = 0$. Substituting the point $D = (a, 0, b)$ into the equation gives $-b^{3}a + (au + bw)(a + b) = 0$ $\Rightarrow au + bw = \frac{b^{3}a}{a + b}$. $\Rightarrow u = \frac{b^{3}}{a + b} - \frac{b}{a}w$. Now, substituting $I' = (ab(a + b), -b^{3}, b(3b^{2} - a^{2}))$ into the equation, we get $-a^{2}(-b^{3})(b(3b^{2} - a^{2})) - b^{2}(b(3b^{2} - a^{2}))(ab(a + b)) - b^{2}(ab(a + b))(-b^{3}) + (u(ab(a + b)) + w(b(3b^{2} - a^{2})))(b^{2}(2b + a)) = 0$ $\Rightarrow u(ab(a + b)) + w(b(3b^{2} - a^{2})))(b^{2}(2b + a) = \frac{ab^{2}(a + b)(3b^{2} - a^{2}) - ab^{4}(a + b) - a^{2}b^{2}(3b^{2} - a^{2})}{2b + a} = \frac{ab^{3}(a + 2b)(b - a)}{2b + a} = ab^{3}(b - a)$. Now, substituting $u = \frac{b^{3}}{a + b} - \frac{b}{a}w$ into the equation above and simplifying it, we get $ab^{4} - b^{2}(a + b)w + w(b(3b^{2} - a^{2})) = ab^{3}(b - a)$ $\Rightarrow ab^{3} - b(a + b)w + w(3b^{2} - a^{2}) = ab^{2}(b - a)$ $\Rightarrow (2b^{2} - a^{2} - ab)w = -a^{2}b^{2}$ $\Rightarrow w = \frac{a^{2}b^{2}}{(a + 2b)(a - b)}$ (Note that $a \neq b$ implies that we did not divide by zero) Now, it remains to substitute $E = (a(-a^{2} - ab + 4b^{2}), 2b^{3}, 2b^{3})$ and verify it lies on $\omega$. $-a^{2}(2b^{3})(2b^{3}) - 2[b^{2}(2b^{3})a(-a^{2} + 4b^{2} - ab)] + (ua(-a^{2} - ab + 4b^{2}) + w(2b^{3}))(a + b)(4b^{2} - a^{2}) = 0$ $\iff u(-a^{3} - a^{2}b + 4ab^{2}) + w(2b^{3}) = \frac{4a^{2}b^{6} + 4ab^{5}(-a^{2} - ab + 4b^{2})}{(a + b)(4b^{2} - a^{2})} = \frac{4ab^{5}(2b - a)(2b + a)}{(a + b)(2b + a)(2b - a)} = \frac{4ab^{5}}{a + b}$, where $a \neq 2b$ because of triangle inequality. However, $u = \frac{b^{3}}{a + b} - \frac{b}{a}w$, so the above equation is equivalent to $\left(\frac{b^{3}}{a + b} - \frac{b}{a}w\right)[a(-a^{2} - ab + 4b^{2})] + w(2b^{3}) = \frac{4ab^{5}}{a + b}$ $\iff \frac{ab^{3}(-a^{2} - ab + 4b^{2})}{a + b} - b(-a^{2} - ab + 4b^{2})w + 2b^{3}w = \frac{4ab^{5}}{a + b}$ $\iff \frac{ab^{3}(-a^{2} - ab + 4b^{2}) - 4ab^{5}}{a + b} = b(a+2b)(b-a)w$ $\iff -a^{2}b^{3} = b(a + 2b)(b - a)w$ Since $w = \frac{a^{2}b^{2}}{(a + 2b)(a - b)}$, the last equality is true, and thus $E$ lies on $\omega$, which shows that $B, D, E, I'$ are concyclic.
20.07.2017 04:52
It's amusing how obvious the complex "bash" is after inversion at $C$. cjquines0 wrote: Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$. Let $M$ be the midpoint of $AC$. Consider the following Claim: In right-triangle $CAM$ ($\angle A=90^{\circ}$), $B$ is the midpoint of $CM$, $I$ is the midpoint of arc $AM$ not containing $C$, $J$ is the reflection of $I$ in $CA$, $D$ lies on ray $CA$ such that $C, B, I, D$ are concyclic, $E$ is the other intersection of $DM$ with circle $\odot(CM)$. Then, $E$ is the reflection of $C$ in $BJ$. (Proof) To see this, note that $BC=BE$ and $$\angle CAB= \angle BCA=\angle MBI=\angle IDC$$so $ABID$ is a parallelogram. Assign complex numbers $B=0$, $C=-1$ and $M=1$, $A=a^2$, $I=a$. By reflection formula, $J=a^2+a-1$ so $$J-B=(a^2+a-1)=A+I-B-M=D-M,$$so $BJ \parallel DM \Longrightarrow BJ \perp CE$. $\square$ Inverting at $C$, the claim reverts to $E$ being the circumcenter of $\triangle BJC$ in the original configuration. As $EB=EJ$; $DE$ externally bisects $\angle BDJ$, so $E$ lies on $\odot(BDJ)$. $\blacksquare$
20.07.2017 08:38
Let $I'$ be the reflection of $I$ in $AC.$ It is enough to show that $\triangle I'IB \sim \triangle I'DE$, since then we will have $\measuredangle I'BD = \measuredangle I'BI = \measuredangle I'ED.$ Since $DE \parallel II'$, reflecting in $AC$ shows that $\measuredangle I'DE = \measuredangle I'IB.$ By SAS similarity, it remains to prove that $\frac{I'I}{IB} = \frac{I'D}{DE}.$ To do so, we make a few observations. 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20.07.2017 10:45
Solution: Let $I^*$ be the reflection of $I $ in $AC $. Also, let $I^*C\cap AE = J $. Claim: $BEDI^*J $ is cyclic. Proof of the claim: We have, $\angle I^*CD = \angle DBC\implies CJ $ is tangent to $\odot (BDC)\implies \angle BCJ = \angle BDC\implies \angle BJC + \angle BDI^* = 180$°. Thus, $BDI^*J$ is cyclic ..... ($1$) Also, $AJ $ bisects $\angle BJC $ and $ED $ bisects $\angle BDI^*$ externally. So, $\angle BDE = \angle BJE\implies BEDJ $ is cyclic ..... ($2$) ($1$) & ($2$) implies our claim which, in turn, completes our proof.
21.07.2017 18:27
This is purely elementary. Since, noone posted a similar solution I am posting the one. Let \(J\) be the reflection of \(I\) in the line \(AC\) and \(AI \cap BC = P.\) In \(\Delta BIP,\) we have: \(\dfrac{IB}{IP} = \dfrac{sin90^{\circ}}{sin \angle IBP}\) \(\dfrac{IB}{2 \times IP} = \dfrac{1}{2 \times sin \angle 45 - A/4}\) \(\dfrac{IB}{IJ} = \dfrac{1}{2 \times sin[ \angle 45 - A/4]} --1\) In \(\Delta EDI,\) we have:: \(\dfrac{DE}{DI} = \dfrac{DE}{DJ} = \dfrac{sin 45 + A/4}{sin 90+A/2} = \dfrac{sin 45 + A/4}{sin 2 \times 45 + A/4} = \dfrac{sin 45 + A/4}{2 \times [sin 45 + A/4][cos 45 +A/4]} = \dfrac{1}{2 \times [sin 45 - A/4]}\) \(--2\) Now, angle chasing is telling us that \(\angle BIJ = EDJ\) Therefore, from \([1] , [2]\) we get \(\Delta BIJ \sim \Delta EDI => \angle JBI = \angle JBD = \angle JED => BEDI\) is cyclic. We can also backup the whole trig part through some constructions like producing side \(IP\) to increase its length by a factor of \(2\) and also reflecting \(DE\) to a point on \(AP\) such that \(DE' = DE\). Now prove the two new triangles similar.
22.07.2017 12:52
Also German TSTST #6.
23.07.2017 15:05
This problem is true for any point on circumcircle of triangle $IBC$ as following Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. Point $P$ chosen on the circumcircle $\omega $ of triangle $IBC$. The line $BP$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $P$ in $AC$ lies on the circumcircle of triangle $BDE$.
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26.07.2017 08:55
Let $I'$ be reflection of $I$ about $AC$. Note that $ED\parallel II'\perp AC$, so $\angle (ED,DI')=\angle (II',DI')=\angle(DI,II')=\angle(BI,II')$. So it suffices to prove that $$\frac{DI'}{ED}=\frac{I'I}{IB}$$as this imply $\triangle I'DE \sim \triangle II'B$, then $\angle(I'E,ED)=\angle(I'B,BI)=\angle(I'B,BD)$, then $BEDI'$ is cyclic. Let $K$ be projection of $I$ about $AC$, then $$\frac{DI'}{ED}=\frac{I'I}{IB}\iff \frac{DI}{DE}=2\frac{IK}{IB}=2\frac{IK}{IC} ..... (1) $$Let $\angle IAC=x$, then $$\frac{IK}{IC}=\sin \angle ICK=\sin (45-\frac{x}{2})=\cos (45+\frac{x}{2})$$$$\frac{DI}{DE}=\frac{\sin \angle IED}{\sin \angle DIE}=\frac{\sin(90+x)}{\sin(x+(45-\frac{x}{2}))}=\frac{\sin(2(45+\frac{x}{2}))}{\sin(45+\frac{x}{2})}=2\cos(45+\frac{x}{2})=2\frac{IK}{IC}$$ So $(1)$ is proven and we are done. Q.E.D
30.07.2017 04:57
LeVietAn wrote: This problem is true for any point on circumcircle of triangle $IBC$ as following Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. Point $P$ chosen on the circumcircle $\omega $ of triangle $IBC$. The line $BP$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $P$ in $AC$ lies on the circumcircle of triangle $BDE$. Generalization Let $\Delta ABC$ be a triangle. Let $\omega$ denote a circle pass through $B,C$ and tangent to $AC$. Let $P$ be any point on $\omega$. Let $BP\cap AC=D$ and $E$ be a point such that $DE\perp AC$ and $BE=CE$. Prove that the reflection of $P$ through $AC$ lies on the circumcircle of $\Delta BDE$.
30.07.2017 05:35
MarkBcc168 wrote: Generalization Let $\Delta ABC$ be a triangle. Let $\omega$ denote a circle pass through $B,C$ and tangent to $AC$. Let $P$ be any point on $\omega$. Let $BP\cap AC=D$ and $E$ be a point such that $DE\perp AC$ and $BE=CE$. Prove that the reflection of $P$ through $AC$ lies on the circumcircle of $\Delta BDE$. Let $ T $ be the reflection of $ P $ in $ AC. $ From $ DB \cdot DT $ $ = $ $ DB \cdot DP $ $ = $ $ {DC}^2 $ we get $ \triangle DBC $ $ \stackrel{+}{\sim} $ $ \triangle DCT, $ so $ D $ is the projection of the circumcenter of $ \triangle BCT $ on its C-symmedian, hence $ E $ is the circumcenter of $ \triangle BCT $ and $ T $ $ \in $ $ \odot (BDE). $
08.08.2017 04:42
Let $I'$ be the reflection of $I$ in $AC$ so $DE$ is the external bisector of $\angle IDI'$ -- it is enough to show that $I'BE$ is isoceles. Let $\theta = \angle IBF$ so we have $\frac{II'}{IB}=2\sin \theta$ by $IF=\frac{II'}{2}$. LoS in $\triangle IDE$ gives $\frac{I'D}{DE}=\frac{ID}{DE}=\frac{\sin 2\theta}{\sin (90^{\circ}+\theta)}=2\sin \theta$, and $\angle IDE=\angle I'ID$ is clear so $\triangle I'DE\sim \triangle I'IB$ and we are done by spiral similarity.
16.09.2017 02:11
23.12.2017 12:31
Let $I'$ be the reflection of $I$ over $AC$. Lemma: $EB = EI'$ Proof: Let $L$ be the reflection of $D$ over $C$. As $E$ lies on the perpendicular bisector of $BC$, the lemma is equivalent to proving that the circumcenter of $\omega_{BCI'}$ is perpendicular to $D$, or equivalently, $B,C,I', L$ are concylic. This, of course, is bary We easily get $I' = (a(a+b) : -b^2: 3b^2-a^2)$ and $L = (2a: 0: b-a)$. Circumcircle of $\omega_{BCL}$ has the equation $ -\sum_{\sigma}a^2yz + (b^2\frac{(b-a)}{b+a} x)(x+y+z) = 0$. So we need to prove that $I'$ lies on this circle, but $$ -a^2b^2(3b^2-a^2) - b^2a(a+b)(2b^2-a^2) + (b-a)b^3a(2b+a) = 0$$, hence this lemma is proved $\box$ Now note that as $ED \perp AC$, easy angle chasing gives $DE$ is the external bisector of $\angle BDI'$. So now we finish the proof by copy pasting the last line of this , and replacing T, O, Q, P by E, D, I', B, we're done.
02.09.2023 05:45
I present the longest synthetic solution thus far. Please do not worry as to how this came about. We define an onslaught of new points. Let $X$ be the intersection of $CI$ with $(ABC)$. Let $K$ be the midpoint of $BI$. Let $L$ be the midpoint of $BC$. Let $M$ be the intersection of $AI$ with $(ABC)$. Let $R$ be the intersection of the perpendicular bisector of $BI$ with $AC$. Let $P$ be the intersection of $(BMR)$ with $AI$. Note by incenter excenter we have that $XBI$ is isosceles. Moreover, $\measuredangle BXI = \measuredangle BXC = \measuredangle BAC$, whence $B$ is the center of the spiral similarity taking $XI$ to $AC$. Moreover, note that $XM$ is the perpendicular bisector of $BI$, whence $X$, $M$, $R$ collinear. Now, \[ \providecommand{\dang}{\measuredangle} \dang PBR - \dang PMR - \dang AMX = \dang ACX = \dang ACI = \dang CBI, \]so $\angle PBC = \angle RBI$. This implies that $B$ is the center of the spiral similarity sending quadrilateral $BXRI$ to $BAPC$. As such it also sends $K$ to $L$. Now by spiral sim, \[ \frac{AL}{AP} = \frac{XK}{XR}, \]and noting that $BD$ and $AX$ are both perpendicular to $XM$, we have \[ \frac{AL}{AP} = \frac{AD}{AR},\]and since $LCDE$ is cyclic, $\frac{AL}{AD} = \frac{AC}{AE}$, whence \[ \frac{AC}{AE} = \frac{AP}{AR} \implies AC\cdot AR = AE \cdot AP, \]whence $ECRP$ is cyclic. Now \[ \providecommand{\dang}{\measuredangle} \dang DER = \dang DEP + \dang PER = \dang DEA + \dang PCR = \dang ACL + \dang PCA = \dang XIB + \dang RIX = \dang RIB = \dang DRB, \]so $BDRE$ is cyclic. At last, since $\angle DIR = 180^\circ - \angle DBR$, it follows that $DBRI'$ is cyclic, whence $BDEI'$ is cyclic. $\textbf{Remark.}$ It appears I have forgotten how to angle chase.
09.09.2023 03:01
I think this is by far the cleanest complex bash in this thread. Set $(ABC)$ as the unit circle with $a=1,b=x^2,c=\tfrac{1}{x^2}$, and let the midpoints of minor arcs $AB$ and $AC$ be $x$ and $\tfrac{1}{x}$ respectively. Then $$d=\frac{x(1+\frac{1}{x^2})-\frac{1}{x^2}(x^2+\frac{1}{x})}{x-\frac{1}{x^2}}=\frac{x^3+x+1}{x(x^2+x+1)}.$$$E=e$ lies on the real axis, hence we have $$\frac{1}{2}\left(1+\frac{1}{x^2}+e-\frac{1}{x^2}e\right)=\frac{x^3+x+1}{x(x^2+x+1)} \implies e(x^2-1)=\frac{2x^4+2x^2+2x}{x^2+x+1}-(x^2+1)=\frac{(x-1)(x^3+1)}{x^2+x+1} \implies e=\frac{x^2-x+1}{x^2+x+1}.$$We also have $I=x+\tfrac{1}{x}-1$, which also lies on the real axis. Hence its reflection $I'$ over $\overline{AC}$ is just $$1+\frac{1}{x^2}-\frac{1}{x^2}\left(x+\frac{1}{x}-1\right)=\frac{x^3-x^2+2x-1}{x^3}.$$We now do a tiny bit of angle chasing: $$\measuredangle BDI'=2\measuredangle BDA=2(\measuredangle DAB+\measuredangle ABD)=3\measuredangle BCA.$$This last angle is equal to the argument of $x^3$, hence it suffices to show that $$\mathbb{R} \ni \frac{\frac{d-e}{i'-e}}{x^3}=\frac{\frac{x^2-\frac{x^2-x+1}{x^2+x+1}}{\frac{x^3-x^2+2x-1}{x^3}-\frac{x^2-x+1}{x^2+x+1}}}{x^3}=\frac{x^2(x^2+x+1)-(x^2-x+1)}{(x^3-x^2+2x-1)(x^2+x+1)-x^3(x^2-x+1)}=\frac{x^4+x^3+x-1}{x^4+x^3+x-1}=1,$$which is clearly true. $\blacksquare$ Remark: If $O$ is the center of $(ABC)$ and $M$ is the midpoint of minor arc $AC$, this solution in fact shows that $\triangle BEI' \stackrel{+}{\sim} \triangle BOF$.
21.11.2023 17:20
length chase/bash
Let $\measuredangle ACB$ be denoted by $\alpha$ .Let $J$ denote the foot of angle bisector of $C$, and let $F$ be the mdipoint of $BC$. Clearly, $E$ is the miquel point, and there is a spiral similarity centered at $A$ taking $II'$ to $BC$. Ww will prove that $I'$ takes $IB$ to $DE$ Claim 1: $\measuredangle I'IB = \measuredangle I'DE$
So it suffices to prove $\frac{II'}{IB}=\frac{ID}{DE}$ and SAS will do the job for us $II' = \frac{AI \cdot BC}{AB}$ $DE = \frac{AD \cdot FC}{AF}$ $\frac{AI \cdot BC}{AB \cdot IB} = \frac{AF \cdot ID }{AD \cdot BC}$ now apply sine rule to get both are equal to $\frac{\sin(\alpha)}{\sin(90+\frac{\alpha}{2})}$
13.01.2024 19:37
This is so easy. Just similar triangles. Still took 1 hr. Not G4 level. Let $I'$ be the reflection of $I$ across $AC$. Observe $ \frac{ED}{DI'}=\frac{BI}{IX} $ (This can be found out using trig calculations) $\angle BII'=\angle EDI'$ Therefore $\triangle BII' \sim \triangle EDI'$ Therefore by spiral similarity $\triangle BII' $ gets sent to $\triangle EDI'$ We are done
21.04.2024 09:02
Let $I'$ be the reflection of $I$ over $\overline{AC}$; so, we're trying to show $BEDI'$ is cyclic. Let $C'$ be the reflection of $C$ over $D$. Claim: $BCI'C'$ is cyclic with center $E$. Proof: since $I$ lies on median $BD$ of $\triangle BC'C$ and $(BIC)$ is tangent to line $AC$, it follows that $I$ is the $B$-humpty point of $\triangle BC'C$, so $I'$ lies on $(BCC')$. Furthermore, the center of $(BCI'C')$ is the intersection of the perpendicular bisectors of $\overline{BC}$ and $\overline{CC'}$, which is $E$. So, ignoring configuration issues, we have \[\angle BEI' = 360^{\circ} - 2\angle BCI' = \angle BDI',\]where the last step follows from straightforward angle chasing.
17.05.2024 19:52
Once you get that $E$ is the centre of $(BCI')$, the problem is nuked
05.06.2024 02:10
let $I'$ be the reflection of $I$ through $AC$, and let $I'C$ intersect $DE$ at $J$ and $AE$ at $K$ we can easily prove that $BIK=CIK=CII'=CI'I$, so $DII'=DI'I=IKC$, and $EDI'K$ are cyclic angle chasing also yields that $AID=KIC$, so $IED=KCI=KBI$, and $BKED$ are cyclic. thus, $BKI'DE$ are all cyclic
16.06.2024 16:53
let $f$ be the incenter $f=\frac{t^2+\frac1t-t-\frac1{t^2}}{t-\frac1t}=t+\frac1t-1=\frac{t^2-t+1}{t}$ the reflection of $f$ across $AC$ is $$x=\overline{(\frac{f-1}{\frac1{t^2}-1})}(\frac1{t^2}-1)+1=\frac{f-1}{t^2-1}\frac{1-t^2}{t^2}+1=\frac{1-f}{t^2}+1=\frac{t^3-t^2+2t-1}{t^3}$$ $d=\frac{t+\frac1t-1-\frac1{t^3}}{t-\frac1{t^2}}=\frac{t^4-t^3+t^2-1}{t^4-t}=\frac{t^3+t+1}{t^3+t^2+t}$ note that $\frac{e-d}{1-\frac1{t^2}}=\frac{e-\overline{d}}{t^2-1}$ so $$e=\frac{t^2d-\overline{d}}{t^2-1}$$ note that $\overline{d}=\frac{1+t^2+t^3}{1+t+t^2}$ so $$e=\frac{\frac{t^4+t^2+t}{t^2+t+1}-\frac{1+t^2+t^3}{1+t+t^2}}{t^2-1}=\frac{t^4-t^3+t-1}{(t^3-1)(t+1)}=\frac{t^3+1}{(t^2+t+1)(t+1)}=\frac{t^2-t+1}{t^2+t+1}$$ TL;DR $b(t^5+t^4+t^3)=t^7+t^6+t^5$ $d(t^5+t^4+t^3)=t^5+t^3+t^2$ $e(t^5+t^4+t^3)=t^5-t^4+t^3$ $x(t^5+t^4+t^3)=t^5+2t^3+t-1$ we wanna show that $\frac{(b-d)(x-e)}{(x-d)(b-e)}$ is real note that $$\frac{(b-d)(x-e)}{(x-d)(b-e)}=\frac{(t^7+t^6-t^3-t^2)(t^4+t^3+t-1)}{(t^3-t^2+t-1)(t^7+t^6+t^4-t^3)}=\frac{(t^4+2t^3+2t^2+2t+1)}{(t^2+1)t}=\frac{t^2+2t+1}{t}=t+2+\frac1t$$ which is clearly real, as desired
01.07.2024 07:33
Claim: $\overrightarrow{\rm AI}$ and $\overrightarrow{\rm I'C}$ intersect at a point $P$ on $(BDI')$. Proof. Angle chasing. Let $E' = \overrightarrow{\rm AI} \cap (BDI’)$. A little more angle chasing shows that $E'D \perp AC$, implying $E = E'$ as desired.
07.07.2024 06:41
Let $I'$ be the reflection of $I$ over $AC$ and let $AI$ meet $CI'$ at $F.$ Then $\angle BDI'=2\angle BDC=2(180-3\angle DBC)=2(180-\angle I'CB)=2\angle BCF=180-\angle BFI',$ so $B,F,I',D$ concyclic. If $G$ lies on $AC$ with $FG\parallel BC,$ then $\angle BDG=\angle FBC=180-\angle BFG,$ so $B,F,G,D$ concyclic. Now the antipode of $G$ lies on the line through $D$ perpendicular to $AC$ and the line through $F$ perpendicular to $FG,$ so it is $E$ so $B,E,D,I'$ concyclic.
10.10.2024 17:15
21.12.2024 15:52
Posting a solution from old, for storage. We employ complex numbers. Set the incircle be the unit circle. Throughout this solution, let $j$ be the complex number representing the incenter ($j=0$). Then, let $T_A$ , $T_B$ and $T_C$ be the tangency points of the incircle to sides $BC,CA$ and $AB$ respectively. WLOG, rotate so that $t_A=1$. Then, let $t_B=x$ for ease of writing. It follows that $t_C = \frac{1}{x}$. From this it follows that, \[a= \frac{2x}{x^2+1} , b = \frac{2}{x+1} \text{ and } c= \frac{2x}{x+1}\]Next, since $D= \overline{BI} \cap \overline{AC}$ we obtain that, \begin{align*} d &= \frac{(\overline{b}j-b\overline{j})(a-c)-(\overline{a}c-a\overline{c})(b-i)}{(\overline{b}-\overline{i})(a-c)-(\overline{a}-\overline{c})(b-i)}\\ &= \frac{-b(\overline{a}c-a\overline{c})}{\overline{b}(a-c)-b(\overline{a}-\overline{c})}\\ &= \frac{\frac{-2}{x+1}\left(\frac{2x}{x^2+1}\cdot \frac{2x}{x+1}-\frac{2x}{x^2+1}\cdot \frac{2}{x+1}\right)}{\frac{2x}{x+1}\left(\frac{2x}{x^2+1}-\frac{2x}{x+1}\right)-\frac{2}{x+1}\left(\frac{2x}{x^2+1}-\frac{2}{x+1}\right)}\\ &= \frac{\frac{-8x(x-1)}{(x+1)^2(x^2+1)}}{\frac{4x^3(1-x)}{(x+1)^2(x^2+1)}-\frac{4(x-1)}{(x+1)^2(x^2+1)}}\\ &= \frac{-8x(x-1)}{-4(x^3+1)(x-1)}\\ d&= \frac{2x}{x^3+1} \end{align*}Now, we need to compute $e$. Since $ED \perp AC$, $\frac{e-d}{d-a} \in i\mathbb{R}$, \begin{align*} \frac{e-d}{d-a} &= - \overline{\left(\frac{e-d}{d-a}\right)}\\ \frac{e-\frac{2x}{x^3+1}}{\frac{2x}{x^3+1}-\frac{2x}{x^2+1}} &= - \left(\frac{e-\frac{2x^2}{x^3+1}}{\frac{2x^2}{x^3+1}-\frac{2x}{x^2+1}}\right)\\ \frac{\frac{e(x^3+1)-2x}{x^3+1}}{\frac{2x^3-2x^4}{(x^3+1)(x^2+1)}} &= - \left(\frac{\frac{e(x^3+1)-2x^2}{x^3+1}}{\frac{2x^2-2x}{(x^3+1)(x^2+1)}}\right)\\ \frac{e(x^3+1)-2x}{x^2} &= e(x^3+1)-2x^2\\ e &= \frac{2x(x^3-1)}{(x^2-1)(x^3+1)}\\ e &= \frac{2x(x^2+x+1)}{(x+1)(x^3+1)} \end{align*}Now, let the reflection of $I$ across $AC$ be $I_1$ represented by the complex number $j_1$. Then, it is clear that $I_1$ is the reflection of $I$ across $T_B$ since $IT_B \perp AC$. Thus, it follows that $j_1=2x$. Now, note that, if $P= \frac{b-d}{d-j_1} : \frac{e-d}{d-j_1}$, \begin{align*} P&= \frac{\frac{2}{x-1}-\frac{2x}{x^3+1}}{\frac{2}{x+1}-2x} \cdot \frac{\frac{2x(x^2+x+1)}{(x+1)(x^3+1)}-2x}{\frac{2x(x^2+x+1)}{(x+1)(x^3+1)}-2x}\\ &= \frac{\frac{2(x^3-x^2-x+1)}{(x+1)(x^3+1)}}{\frac{-2(x^2+x-1)}{x+1}} \cdot \frac{\frac{-2x^3(x^2+x-1)}{(x^3+1)(x+1)}}{\frac{2x^3}{(x+1)(x^3+1)}}\\ &= \frac{x^3-x^2-x+1}{(x^3+1)(x^2+x-1)} \cdot (x^2+x-1)\\ &= \frac{x^3-x^2-x+1}{x^3+1} \end{align*}Now, note that \[\overline{P}=\overline{\left(\frac{x^3-x^2-x+1}{x^3+1}\right)} = \frac{\frac{1}{x^3}-\frac{1}{x^2}-\frac{1}{x}+1}{\frac{1}{x^3}+1}=\frac{x^3-x^2-x+1}{x^3+1}=P\]which proves that $\frac{b-d}{d-j_1} : \frac{e-d}{d-j_1} \in \mathbb{R}$ as desired. Thus, points $B,D,E$ and $I_1$ must be concyclic which finishes the problem.
21.12.2024 18:33
Let $I'$ be the reflection of $I$ in $AC$. Then notice that $BI$ and the $CI'$ intersect on the circumcircle call the point $M$. Consider the inversion centered $M$ with radius $MI$. It's easy to see our inversion sends $I'$ to the intersection of $AI$ and $CI'$ call that point $I"$ and it sends $B$ to $D$. So proving $BEDI"$ is cyclic will be enough. Note that $MI^2=MI' \cdot MI"$ so $\angle EI"B=\angle II"B=\angle II"C=\angle II"I'=\angle MII'=\angle DII'=\angle EDI=\angle EDB$ done.
31.12.2024 15:01
Let $I'$ be the reflection of $I$ over $AC$ and let $AI\cap CI'=K$. \begin{align*} \measuredangle BKE &= \measuredangle CKE \\&= \measuredangle KAC + \measuredangle ACK\\&= 90^{\circ}+ \measuredangle BCD+ \measuredangle DCI'\\&=90^{\circ} + \measuredangle BCD+\measuredangle DBC\\& =90^{\circ}+ \measuredangle BDC\\&= \measuredangle BDE\implies \boxed{K\in \odot(BED)} \end{align*}\begin{align*} \measuredangle BKI'&= 2\measuredangle BKE\\& =2\measuredangle BDE\\& =2(90^{\circ}+\measuredangle BDC)\\&= \measuredangle BDI'\implies \boxed{I'\in \odot(BEDK)} \end{align*}
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21.01.2025 10:39
Here is a solution with Ptolemy's sine lemma, communicated to me by trigadd123. Set $\angle BAC =2\alpha$. By Ptolemy's sine lemma, we wish to prove that $$DE\cdot\sin(\angle I'DB)+DI'\cdot\sin(\angle EDB) = DB\cdot\sin(\angle EDI').$$We quickly notice that $DI'=DI$, $\angle I'DB=90^{\circ}+ 3\alpha$ and $\angle EDB = 180^{\circ}-\angle EDI' = 45^{\circ}-\frac{3\alpha}{2}$. After plugging in, we are left with showing that $$BI\sin\left(45^{\circ} - \frac{3\alpha}{2}\right)=DE\cos (3\alpha).$$However, by the law of sines in $\triangle ABI$ and $\triangle ADI$, we get $$BI=AI\cdot\frac{\sin\alpha}{\sin(45^{\circ}-\frac{\alpha}{2})}=AD\cdot\frac{\sin(45^{\circ}+\frac{3\alpha}{2})\sin\alpha}{\sin(45^{\circ}+\frac{\alpha}{2})\sin(45^{\circ}-\frac{\alpha}{2})},$$and clearly $DE=AD\tan\alpha$. We are thus left with proving that $$\frac{\sin(45^{\circ}+\frac{3\alpha}{2})\sin(45^{\circ}-\frac{3\alpha}{2})}{\sin(45^{\circ}+\frac{\alpha}{2})\sin(45^{\circ}-\frac{\alpha}{2})}=\frac{\cos 3\alpha}{\cos\alpha},$$which is clearly true, as the numerator and denominator of the LHS are respectively $\frac{\cos 3\alpha}{2}$ and $\frac{\cos\alpha}{2}$.