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$D,E,F$ are, in fact, the reflections of $O$ (the circumcenter of $ABC$) in $CA,AB,BC$ respectively. This means that $BC=DE,\ BC\|DE$, and the same holds for the other pairs of corresponding sides of $ABC, D EF$. This means that $BCDE,CAEF,ABFD$ are parallelograms, so $N$, the persepctor of $ABC$ and $D EF$, is the common midpoint of $AF,BD,CE$. We thus have $\sum\frac{DB}{DD'}=2\sum\frac{DN}{DD'}=2(3-\sum\frac{ND'}{DD'})$, which is equal to $2$ times $3-\left(\frac{S(ENF)}{S(D E F)}+\frac{S(NFD)}{S(D E F)}+\frac{S(DNE)}{S(D E F)}\right)$, so the result is $4$.

I get the feeling that there is something seriously wrong with the solution Grobber gave, although the answer is 4 indeed...

You are right, Grobber's solution is incorrect. He probably got confused by the asymmetric notations (this also happened to me when I saw the problem first). However, the problem is still quite easy: In the following, $\left|P_1P_2P_3\right|$ will mean the (non-directed) area of a triangle $P_1P_2P_3$. Let d be the circle with center D and radius DA = DC. This circle d clearly passes through the points A and C. Similarly, we have a circle e with center E and radius EA = EB, which passes through the points A and B, and a circle f with center F and radius FB = FC, which passes through the points B and C. Let the two circles d and e intersect at a point S (apart from A). In the circle e, the angle < BEA is the central angle of the chord AB, while the angle < ASB is the obtuse chordal angle over this chord; hence, $\measuredangle BEA=2\cdot\left(180^{\circ}-\measuredangle ASB\right)$. Comparing this with $\measuredangle BEA=2\cdot\measuredangle ABC$, we get < ABC = 180° - < ASB, so that < ASB = 180° - < ABC. Similarly, we find < ASC = 180° - < BAC, since the point S lies on the circle d. Hence, < BSC = 360° - < ASB - < ASC = 360° - (180° - < ABC) - (180° - < BAC) = < ABC + < BAC = 180° - < ACB (by the sum of angles in triangle ABC). Thus, < ACB = 180° - < BSC. Hence, $\measuredangle CFB=2\cdot\measuredangle ACB$ becomes $\measuredangle CFB=2\cdot\left(180^{\circ}-\measuredangle BSC\right)$. Now, since the points B and C lie on the circle f, and the angle < CFB is the central angle of the chord BC in this circle, this equation yields that < BSC is the obtuse chordal angle over the chord BC in this circle f; thus, the point S lies on the circle f. Hence, the point S lies on all three circles d, e, f. The points B and S, being the two common points of the circles e and f, must be symmetric to each other with respect to the line EF joining the centers E and F of these circles; hence, the triangles BEF and SEF are congruent, so that | BEF | = | SEF |. The two triangles DD'E and BD'E have a common altitude from the vertex E; thus, their areas are proportional to their bases DD' and BD'. In other words, $\frac{\left|DD^{\prime}E\right|}{\left|BD^{\prime}E\right|}=\frac{DD^{\prime}}{BD^{\prime}}$. In other words, $BD^{\prime}\cdot\left|DD^{\prime}E\right|=DD^{\prime}\cdot\left|BD^{\prime}E\right|$. Similarly, $BD^{\prime}\cdot\left|DD^{\prime}F\right|=DD^{\prime}\cdot\left|BD^{\prime}F\right|$. Hence, $BD^{\prime}\cdot\left|DEF\right|=BD^{\prime}\cdot\left(\left|DD^{\prime}E\right|+\left|DD^{\prime}F\right|\right)=BD^{\prime}\cdot\left|DD^{\prime}E\right|+BD^{\prime}\cdot\left|DD^{\prime}F\right|$ $=DD^{\prime}\cdot\left|BD^{\prime}E\right|+DD^{\prime}\cdot\left|BD^{\prime}F\right|=DD^{\prime}\cdot\left(\left|BD^{\prime}E\right|+\left|BD^{\prime}F\right|\right)=DD^{\prime}\cdot\left|BEF\right|$ $=DD^{\prime}\cdot\left|SEF\right|$, so that $\frac{BD^{\prime}}{DD^{\prime}}=\frac{\left|SEF\right|}{\left|DEF\right|}$. Hence, $\frac{DB}{DD^{\prime}}=\frac{DD^{\prime}+BD^{\prime}}{DD^{\prime}}=1+\frac{BD^{\prime}}{DD^{\prime}}=1+\frac{\left|SEF\right|}{\left|DEF\right|}$. Similarly, $\frac{EC}{EE^{\prime}}=1+\frac{\left|SFD\right|}{\left|DEF\right|}$; $\frac{FA}{FF^{\prime}}=1+\frac{\left|SDE\right|}{\left|DEF\right|}$. Thus, $\frac{DB}{DD^{\prime}}+\frac{EC}{EE^{\prime}}+\frac{FA}{FF^{\prime}}=\left(1+\frac{\left|SEF\right|}{\left|DEF\right|}\right)+\left(1+\frac{\left|SFD\right|}{\left|DEF\right|}\right)+\left(1+\frac{\left|SDE\right|}{\left|DEF\right|}\right)$ $=3+\frac{\left|SEF\right|+\left|SFD\right|+\left|SDE\right|}{\left|DEF\right|}=3+\frac{\left|DEF\right|}{\left|DEF\right|}=3+1=4$. And the problem is solved. Darij

Let $AD \cap CE = A'$, $DA \cap FB = B'$, and $FB \cap EC = C'$. As $\angle B'AB = 90^{\circ}$ and $FB = FA$, $F$ is the circumcenter of $\triangle ABB'$. Similarly, $E$ is the circumcenter of $\triangle BCC'$ and $D$ is the circumcenter of $\triangle CAA'$. By Miquel's theorem on $\triangle A'B'C'$ with respect to $A$, $B$, and $C$, the circumcircles of $\triangle ABB'$, $\triangle BCC'$, and $\triangle CAA'$ pass through a common point $M$. Any two circles are symmetric about the lines joining their centers, so $[EBF] = [FME]$, $[CDF] = [DMF]$, and $[AED] = [EMD]$. Now, \begin{align*} \frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'} &= 3 + \frac{BD'}{D'D} + \frac{CE'}{E'E} + \frac{AF'}{F'F} \\ &= 3 + \frac{[FEB]}{[FDE]} + \frac{[CDF]}{[DEF]} + \frac{[AED]}{[EFD]} \\ &= 3 + \frac{[FME] + [DMF] + [EMD]}{[EFD]} \\ &= 3 + \frac{[EFD]}{[EFD]} \\ &= 4. \end{align*}

Let the circumcenter be $O$, and WLOG $OA=OB=OC=1$. Now, angle chasing with law of sines yields that $[AED]=\frac {sin(b)sin(c)} {2sin(a)}$ and $[EOD]=\frac {sin(c)sin(a)} {2sin(b)}$. So it follows that $[AED]+[DCF]+[BEF]=[DEF]$. Now, note that $\frac {DB} {DD'}=1+\frac {[BEF]} {[DEF]}$, so the cyclic sum of this is $4$ and we are done.

Solution 1: Let $R$ be the circumradius of the triangle and let $K$ be its area. Note that $BE = \frac{a}{2\sin C} = R\frac{a}{c}$, and $\angle FBE=180^\circ -\angle C$, so \[ [BEF] = \frac{1}{2} \left(R\frac{a}{c}\right)\left(R\frac{c}{b}\right)(\sin C)=\frac{R}{4}\cdot \frac{ac}{b}. \]Additionally, it's easy to see \[ [BFC] = \frac{1}{4} a^2 \cot C =\frac{R}{2}\cdot \frac{a^2\cos C}{c} =\frac{R}{4}\cdot \frac{a^4+a^2b^2-a^2c^2}{abc} \]Lemma: \[ 2([ADE]+[BEF]+[CFD])=[ABC]+[BCF]+[CAD]+[ABE] \]Proof: The left hand side is \[ \frac{R}{2abc}(a^2b^2+b^2c^2+c^2a^2)=\frac{1}{8K}\cdot (a^2b^2+b^2c^2+c^2a^2) \]The right hand side is \[ K+\frac{R}{4}\cdot \frac{a^4+b^4+c^4}{abc}=K+\frac{1}{16K}\cdot (a^4+b^4+c^4) \]so their equality follows from \[ 16K^2= 2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4. \] Now, from the lemma it follows that \[ [ADE]+[BEF]+[CFD]=[DEF] \]but \[ \frac{DB}{DD'} = 1+\frac{D'B}{DD'} = 1+\frac{[BEF]}{[DEF]} \]so \[ \frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}=4. \]Solution 2: Consider the circle $\Gamma_D$ centered at $D$ passing through $A$ and $C$; it is tangent to $AB$ at $A$ since $DA\perp AB$. Then $\Gamma_D,\Gamma_E,$ and $\Gamma_F$ concur at the Brocard point $\Omega$. Then $\Omega$ is the reflection of $B$ over $EF$, implying \[ [ADE]+[BEF]+[CFD]=[\Omega DE] +[\Omega EF] +[\Omega FD]=[DEF] \]and we can finish as the last solution.

The desired sum is 4. We first translate the lengths into areas: \[\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}= \frac{[BFDE]}{[DEF]} + \frac{[CDEF]}{[DEF]} + \frac{[AEFD]}{[DEF]} = 3 + \frac{[ADE] + [BEF] + [CFD]}{[DEF]}.\] Let $O$ be the circumcenter of $ABC$ and $R$ the circumradius. Since $\triangle AEO\sim\triangle ABC$, we have $EA = EB = R\cdot \frac{c}{b}$. Similarly, $FB = FC = R\cdot\frac{a}{c}$ and $DC = DA = R\cdot\frac{b}{a}$. Since $\angle EAD = 180 - B$, by LoC on $\triangle ADE$, \begin{align*} DE^2 &= R^2\left(\frac{b^2}{a^2} + \frac{c^2}{b^2} - 2\frac{c}{a}\cos (180 - B)\right)\\ &= R^2\left(\frac{b^2}{a^2} + \frac{c^2}{b^2} +2\frac{c}{a}\cdot\frac{a^2+c^2-b^2}{2ac}\right)\\ &= R^2\left(\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2}\right). \end{align*}As a result, $DE : EF : FD = c : a : b$. Then $\angle DEF = \angle B = \frac{1}{2}\angle AEB$, and since $EA = EB$, this means the reflection of $A$ over $DE$ is the reflection of $B$ over $EF$, and by symmetry is also the reflection of $C$ over $FD$. So $[ADE] + [BEF] + [CFD] = [DEF]$, giving the original sum a value of 4.

I have a very clean solution:

Let $O$ be the circumcenter. Note that $\tfrac{DB}{DD'}=\tfrac{[DEBF]}{[DEF]}$, so \[\frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}=\frac{2[DEF]+[AEBFCD]}{[DEF]}\]Since $D,E,F$ are reflections of $O$ across the sides of $\ \triangle ABC$, $[AEBFCD]=2[ABC].$ We claim that the answer is $4$, which results from $[DEF]=[ABC]$. We have that $EBOA$, $FCOB$, and $DAOC$ are rhombi. Therefore, $AE\parallel BO\parallel FC$ and $AE=AO=BO=FC$, which implies that $AEFC$, $EBCD$, and $BFDA$ are parallelograms. Thus, the midpoints of $AF$, $BD$, and $CE$ coincide, at say $P$. Thus, $A$ and $F$, $B$ and $E$, and $C$ and $F$ are all reflections of each other about $P$. In particular, $\triangle DEF\cong \triangle ABC$. We are done.

@above is faulty, misread the angle conditions (same mistake as what post #3 made) this sol is similar to anser's above, but with a tiny bit more computation at the end instead of the reflection finish The answer is $4$. Let $O$ denote the circumcenter, and WLOG the circumradius $R$ is $1$. Claim: $[ADE] + [BEF] + [CFD] = [DEF]$. Proof. Length bash Note that the angle conditions translate to $\triangle AEB \sim \triangle AOC$ and $\triangle AOB \sim \triangle ADC$, so we have $AE = R \cdot \frac{c}{b} = \frac{c}{b}$ and $AD = \frac{b}{a}$ - also note that $\angle EAD = 90 - \angle A + 90 - \angle B + A = 180 - \angle B$, so we may now use LOC to determine $DE$; \begin{align*} DE^2 &= \frac{b^2}{a^2}+\frac{c^2}{b^2}-2\frac ca \cos(180-B) \\ &= \frac{b^2}{a^2} + \frac{c^2}{b^2} +2\frac{c}{a}\cdot\frac{a^2+c^2-b^2}{2ac} \\ &= \frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2}. \end{align*} However, note that this implies that $\triangle DEF \sim \triangle ABC$ (ratio of corresponding sides is the same), so the ratio between their areas is equal to $\frac{DE^2}{c^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}$. Now we may compute $[DEF]$ to be equal to (using the area formula $[ABC] = \frac{abc}{4R} = \frac{abc}4$) $\frac{a^2b^2+b^2c^2+c^2a^2}{4abc}$. We now find an expression for $[ADE]$. However, this is easy - just use the sine area formula, so \begin{align*} [ADE] &= \frac12 AE \cdot AD \cdot \sin(B) \\ &= \frac{c}{a} \cdot \frac b4 = \frac{bc}{4a} \end{align*} Hence we must show that $\frac{bc}{4a} + \frac{ca}{4b} + \frac{ab}{4c} = \frac{a^2b^2+b^2c^2+c^2a^2}{4abc}$, which is true by inspection. $\blacksquare$ To finish, note that \[\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}= \frac{[BFDE]}{[DEF]} + \frac{[CDEF]}{[DEF]} + \frac{[AEFD]}{[DEF]} = 3 + \frac{[ADE] + [BEF] + [CFD]}{[DEF]} = 4\]and we are done. $\square$