Coordinates too imba!

For those of us that can't recall much about relationships between perpendicular bisectors, angle bisectors, and circumcircles. Here is a straighforward bash (albeit a bit long).

I made a diagram (sorry it's a picture, not Asymptote)

Let G be midpoint of arc BAC and H be intersection point of AG and BC. HBDC is harmonic division, so GA,GB,GD,GC is. And since $\angle GFE=\angle DFE=90$ , GA,GB,GF,GC is harmonic, and thus ABFC is harmonic, which says AF is symmedian. Invert all the figure by A and radius $\sqrt{AB \times AC}=\sqrt{15}$. Then F becomes midpoint of B'C'. By pappus's midline theorem $AF'^{2}=\frac{2*3^{2}+2*5^{2}-7^{2}}{4}=\frac{19}{4}$, so $AF^{2}=\frac{\sqrt{15}^{4}}{AF'^{2}}=\frac{900}{19}$. Done.

Apply $\sqrt{bc}$ inversion. We get that $D$ and $E$ are inverses and since $\angle DFE = 90^{\circ}$, we have that $\angle D'F'E' = 90^{\circ}$ and since the circumcircle gets inverted to $B'C'$, we have that $F'$ is the foot of the perpendicular of $D'$ to $B'C'$. Since $D'$ is the midpoint of arc $B'C'$ of circle $A'B'C'$, we have that $F'$ is the midpoint of $B'C'$. We can use Stewart's theorem to find $AF' = \frac{\sqrt{19}}{2}$. We have $AF = \frac{bc}{AF'} = \frac{30}{\sqrt{19}} \implies AF^2 = \frac{900}{19}$.

Is there a solution that uses a homothety?

Oops how do you know that FD is the bisector of CFB? ah @below thx (the solution just said because DFE was right and i was wondering if i was being ignorant/stupid oops)

@above $\angle{DFE}$ is right. Since $E$ is the midpoint of arc BC not containing A, FD must meet the circumcircle at the antipode of E wrt the circumcircle of ABC, which is the midpoint of arc BAC. Since F and A are on opposite sides of BC, this point is also the midpoint of arc BC not containing F, so FD must bisect $\angle{BFC}$

miss the days when aime was actually ok quality Claim: $AF$ is a symmedian. Proof: We invert around $A$ with radius $\sqrt{AB \cdot AC}$ followed by a reflection across the angle bisector of $\angle BAC$. This swaps $B$ and $C$, along with $D$ and $E$. Moreover, $F$ gets sent to a point on line $BC$ such that $\angle EMD = 90$. But this point is clearly the midpoint of $BC$, so $AF$ is a symmedian, as desired. Now, let $M$ be the midpoint of $BC.$ By symmedian properties we know that $\frac{AF}{AB} = \frac{AC}{AM}$, or $AF = \frac{AC \cdot AB}{AM} = \frac{15}{AM}$, so it suffices to compute $AM$. However, it is well-known that $AM = \frac 12 \sqrt{2AB^2+2AC^2-BC^2}$, so a calculation gives $AM = \frac12 \sqrt{19}$, and so $AF^2 = \frac{900}{19} = \boxed{919}$.

Another different solution? Let the $A$-antipode be $A’$. By inscribed angle theorem, $\angle DAF = \angle EA’F$. Note that $\angle DFA = 90 - \angle DFA = \angle EFA’$ giving us that $\triangle DAF \sim \triangle EA’F$. Therefore, $\frac{A’F}{AF} = \frac{A’E}{AD}$. Using Pythag on $\triangle A’AF$, we see that: $$(2R)^2 = AF^2 + A’F^2 = AF^2(1 + \frac{A’E^2}{AD^2}),$$where $R$ is the radius of $w$. We can solve for $AF^2$ once we find the other values.

$= \frac{4}{3}$. Plugging in these values into the equation, we find that $AF^2 = \frac{900}{19} \rightarrow \boxed{919}$.

...since when was AIME #15 trivial by a few olympiad lemmas A problem from Russia 2009 (also in EGMO!) tells us that $AF$ is a symmedian of $\triangle ABC$. EGMO Lemma 4.26 then tells us $\triangle ABF$ and $\triangle AMC$ are similar, where $M$ is the midpoint of $BC$. Observe that $$AM^2 = \frac{50+18-49}4 = \frac{19}4,$$so now $$\frac{AC^2}{AK^2} = \frac{AM^2}{AB^2} \implies \frac 9{AK^2} = \frac{\frac{19}4}{25} \implies AF^2 = \frac{900}{19},$$yielding the answer as $\boxed{919}$.

HamstPan38825 wrote: ...since when was AIME #15 trivial by a few olympiad lemmas A problem from Russia 2009 (also in EGMO!) tells us that $AF$ is a symmedian of $\triangle ABC$. EGMO Lemma 4.26 then tells us $\triangle ABF$ and $\triangle AMC$ are similar, where $M$ is the midpoint of $BC$. Observe that $$AM^2 = \frac{50+18-49}4 = \frac{19}4,$$so now $$\frac{AC^2}{AK^2} = \frac{AM^2}{AB^2} \implies \frac 9{AK^2} = \frac{\frac{19}4}{25} \implies AF^2 = \frac{900}{19},$$yielding the answer as $\boxed{919}$. how to memorize egmo lemmas (including all the problems :rapid:)

At this rate HamstPan will perfect the AIME

samrocksnature wrote: At this rate HamstPan will perfect the AIME ..as if he can't right now

skipped a few (trivial) steps but most ppl should understand this oops

Walk-Through: $\bullet$ Let $M$ be the midpoint of $BC$ and $P$ be the projection of $A$ onto $BC$. $\bullet$ Angle chasing implies that $EF$ passes through the midpoint of arc $BAC$, so $ABFC$ is harmonic, i.e. $AF$ is the $A$-symmedian of $ABC$. $\bullet$ Now, isogonality implies $ABF \sim AMC$. $\bullet$ Heron's implies $[ABC] = \frac{15 \sqrt{3}}{4}$, so $AP = \frac{15 \sqrt{3}}{14}$ and $BP = \frac{65}{14}$. $\bullet$ Thus, we know $MP = BP - BM = \frac{8}{7}$, and Pythag implies $AM = \frac{\sqrt{931}}{14}= \frac{\sqrt{19}}{2}$. $\bullet$ Now, similarity ratios finishes. Remarks: This problem is essentially trivial for anyone who is experienced in Olympiad Geometry, as this definition of the symmedian is well-known. (Also, I just recalled that the median formula exists... .)

Can someone please point out the mistake? Use diagram in post 9. Also just assume that triangle CEB is equilateral (I proved this but prefer not to detail it right now - other posts confirm this result) If the intersection of CE and AF is U, then angle FCU = angle FAE because they subtend the same arc and obviously angle CUF = angle AUE, so triangles CUF and AUE are similar. Hence, CU/AU = FU/EU. But angle FUE = angle CUA, so by SAS triangles CUA and FUE are congruent. Therefore, FE = AC = 3. Next, we have angle CEF = angle CAF, so by SAS triangles CAF and FEC are congruent. This means AF = EC = 7, which is wrong.

What is a symmedian part 2 (I know what it is now I just didn't spot it) One can check that $\angle A=120^\circ$, and since $\angle EDF=90^\circ$ $\overline{DF}$ passes through the arc midpoint of $BC$ that's not $E$. From here miss the fact that this implies $\overline{DF}$ bisects $\angle BFC$ and note that this means $DMEF$ and $ADMT$ are cyclic where $M$ is the midpoint of $\overline{BC}$. Since $AFET$ is cyclic we have $\tfrac{AF}{TE}=\tfrac{DA}{DT}$. A handy formula tells us $AD=\tfrac{15}{8}$ so now we calculate $DT$. Since $BD=\tfrac{35}{8}$ and $BM=\tfrac{7}{2}$ we have $DM=\tfrac{7}{8}$. Further $\triangle CMT$ is 30-60-90 so $MT=\tfrac{CM}{\sqrt{3}}=\tfrac{7}{2\sqrt{3}}$. Pythag gives $DT=\tfrac{7\sqrt{19}}{8\sqrt{3}}$. Thus $\tfrac{DA}{DT}=\tfrac{15\sqrt{3}}{7\sqrt{19}}$. Now $R=\tfrac{abc}{4A}=\tfrac{7}{\sqrt{3}} \implies TE=\tfrac{14}{\sqrt{3}}$, which gives $AF=\tfrac{30}{\sqrt{19}} \implies \boxed{919}$.

OlympusHero wrote: Can someone please point out the mistake? Use diagram in post 9. Also just assume that triangle CEB is equilateral (I proved this but prefer not to detail it right now - other posts confirm this result) If the intersection of CE and AF is U, then angle FCU = angle FAE because they subtend the same arc and obviously angle CUF = angle AUE, so triangles CUF and AUE are similar. Hence, CU/AU = FU/EU. But angle FUE = angle CUA, so by SAS triangles CUA and FUE are congruent. Therefore, FE = AC = 3. Next, we have angle CEF = angle CAF, so by SAS triangles CAF and FEC are congruent. This means AF = EC = 7, which is wrong. Please help with this

Employ Barycentric Coordinates on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. My solution to Russia 2009 tells us that $F=(-a^2:2b^2:2c^2)$. Of course, $$\overrightarrow{AF}=\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1,\frac{2b^2}{-a^2+2b^2+2c^2},\frac{2c^2}{-a^2+2b^2+2c^2}\right)$$so $$|\overrightarrow{AF}|^2=-a^2\left(\frac{2b^2}{-a^2+2b^2+2c^2}\right)\left(\frac{2c^2}{-a^2+2b^2+2c^2}\right)-b^2\left(\frac{2c^2}{-a^2+2b^2+2c^2}\right)\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1\right)-c^2\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1\right)\left(\frac{2b^2}{-a^2+2b^2+2c^2}\right)$$by the barycentric distance formula. Fortunately, the sides of this triangle are small integers and after a bit of computation, we find $AF^2 = \frac{900}{19} \rightarrow \boxed{919}$.

Firstly, law of cosines gives $\angle A = 120^\circ$. Then $\angle ECB = \angle EAB = 60^\circ$, and. $\angle CEB = 180 - 120 = 60^\circ$, so $\triangle CEB$ is equilateral and thus $CB = BE = EC = 7$. By Stewart's theorem, we can find that $AD = \frac{15}{8}$, so $\frac{15}{8}DE = \frac{21}{8} \cdot \frac{35}{8} \implies DE = \frac{49}{8}$, so $DO = EO = \frac{49}{16}$; then finish either with coordinates or just do #6 Also; writing similarity ratios is hard, #30

Copying ikechen's style. Walkthrough $\bullet$ Use cosine's law on $\triangle ABC$ to get that $\angle A = 120$. $\bullet$ Now note that $FD$ bisects $\angle BFC$, since $\angle DFC = 30$. $\bullet$ By angle bisector theorem we have $BF = 3x, CF = 5x$. $\bullet$ Now use cosine's law on $\triangle BFC$ to get the value of $x$. $\bullet$ Now use ptolemy's on $ABFC$ to finish.

unique but lengthy approach?

With the Law of Cosines, we see that $\angle BAC =120^\circ$, hence signalling we could set up coordinates. Let $A=(0,0)$, $B=(5,0)$, and $C = \left(-\frac{3}{2}, \frac{3}{2} \sqrt{3} \right)$. We solve that \[\omega: \ \left(x-\frac{5}{2} \right)^2 + \left(y-\frac{11}{2\sqrt{3}} \right)^2 = \frac{49}{3},\] and \[\overline{BC}: \ y = -\frac{3\sqrt{3}}{13}x +\frac{15\sqrt{3}}{13}\]\[\overline{AE}: \ y = \sqrt{3}x. \] Thus, we get $D=\left(\frac{15}{16}, \frac{15\sqrt{3}}{16} \right)$ and $E=(4,4\sqrt{3}).$ This means that $M$, the center of $\gamma$, has coordinates $\left(\frac{79}{32}, \frac{79\sqrt{3}}{32} \right)$. Hence, \[\gamma: \ \left(x-\frac{79}{32} \right)^2 + \left(y-\frac{79\sqrt{3}}{32} \right)^2 = \frac{2401}{256}\]\[\implies F = \left(\frac{15}{19}, \frac{75\sqrt{3}}{19} \right).\] The desired result is $AF^2 = \frac{900}{19} \implies \boxed{919}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.050628616178989, xmax = 11.046649890111382, ymin = -1.591755275635423, ymax = 8.174727877807914; /* image dimensions */ /* draw figures */ draw((0,0)--(5,0), linewidth(1)); draw((5,0)--(-1.5,2.598076211353316), linewidth(1)); draw((-1.5,2.598076211353316)--(0,0), linewidth(1)); draw(circle((2.5,3.1754264805429417), 4.041451884327381), linewidth(1)); draw((0,0)--(4,6.92820323027551), linewidth(1)); draw(circle((2.46875,4.2760004311856665), 3.0625), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0.14283861267706864,0.06608862466272797), NE * labelscalefactor); dot((5,0),dotstyle); label("$B$", (4.92215580679371,0.12608862466272797), NE * labelscalefactor); dot((-1.5,2.598076211353316),linewidth(4pt) + dotstyle); label("$C$", (-1.4467602938751747,2.6968480279061158), NE * labelscalefactor); dot((0.9375,1.6237976320958225),linewidth(4pt) + dotstyle); label("$D$", (0.9798443764760604,1.6238035808840857), NE * labelscalefactor); dot((4,6.92820323027551),linewidth(4pt) + dotstyle); label("$E$", (4.043132450434302,7.021490014670693), NE * labelscalefactor); dot((2.46875,4.2760004311856665),linewidth(4pt) + dotstyle); label("$M$", (2.517494860659021,4.166640350573549), NE * labelscalefactor); dot((0.789473684210522,6.837042661456093),linewidth(4pt) + dotstyle); label("$F$", (0.7356896435839078,6.937399753816937), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

$AF$ is symmedian and then just appolonious and similarity