Question: Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$

Let the common angle be \(\theta\).

### 1. Place the triangle in coordinates

Set
\[
A=(0,0),\qquad B=(55,0),\qquad C=\left(\frac{55}{2},\frac{55\sqrt3}{2}\right).
\]
Since \(AF=40\), \(CE=30\), and \(BD=7\),
\[
F=(40,0),
\]
\[
E=\frac{25}{55}C=\left(\frac{25}{2},\frac{25\sqrt3}{2}\right),
\]
and
\[
D=B+\frac7{55}(C-B)=\left(\frac{103}{2},\frac{7\sqrt3}{2}\right).
\]

The rays \(FB,DC,EA\) have direction angles \(0^\circ,120^\circ,240^\circ\), respectively. Because the triangle is oriented counterclockwise and \(P\) is inside it, the rays \(FP,DP,EP\) have direction angles
\[
\theta,\qquad \theta+120^\circ,\qquad \theta+240^\circ.
\]

### 2. Write normal vectors to these three lines

Let \(c=\cos\theta\) and \(s=\sin\theta\). A normal vector to a line in direction \(\alpha\) is
\[
n(\alpha)=(-\sin\alpha,\cos\alpha).
\]
Thus the normals to \(FP,DP,EP\) are
\[
n_0=(-s,c),
\]
\[
n_1=\left(\frac{s-\sqrt3c}{2},\frac{-\sqrt3s-c}{2}\right),
\]
\[
n_2=\left(\frac{s+\sqrt3c}{2},\frac{\sqrt3s-c}{2}\right).
\]
These satisfy
\[
n_0+n_1+n_2=0.
\]

If \(P\) lies on all three lines, then
\[
n_0\cdot(P-F)=n_1\cdot(P-D)=n_2\cdot(P-E)=0.
\]
Adding these equations and using \(n_0+n_1+n_2=0\) gives
\[
n_0\cdot F+n_1\cdot D+n_2\cdot E=0.
\]

Now compute:
\[
n_0\cdot F=-40s,
\]
\[
n_1\cdot D
=\frac{s-\sqrt3c}{2}\cdot\frac{103}{2}
+\frac{-\sqrt3s-c}{2}\cdot\frac{7\sqrt3}{2}
=\frac{41s-55\sqrt3c}{2},
\]
and
\[
n_2\cdot E
=\frac{s+\sqrt3c}{2}\cdot\frac{25}{2}
+\frac{\sqrt3s-c}{2}\cdot\frac{25\sqrt3}{2}
=25s.
\]
Therefore
\[
-40s+\frac{41s-55\sqrt3c}{2}+25s=0,
\]
so
\[
11s-55\sqrt3c=0.
\]
Hence
\[
\tan\theta=\frac{s}{c}=5\sqrt3,
\]
and consequently
\[
\tan^2\theta=(5\sqrt3)^2=75.
\]

Final Answer: 75

Explanation:
We need solve. Need step-by-step final last exact. Need deeply reason. Need formulate geometry.

We have equilateral side 55, D on BC BD7, E on CA CE30, F on AB AF40. P inside, angles AEP=BFP=CDP presumably common acute? Need tan².

Need derive.

Set coordinates likely exploit rotations/cyclic? Equal directed angles from points on sides involving P. Three lines EP, FP, DP make certain angles with side rays EA, FB, DC. Need locate P and angle theta satisfying triangle contact distances/rotation.

Let's set coordinate. Equilateral perhaps A top, B,C base. Need directed coordinate.

Let A=(0,? maybe). Choose convenient orientation. Need understand angle AEP: rays EA along CA from E toward A. Since CE=30, AC55, AE=25. FB: F on AB, AF40 so FB15. DC: D on BC, BD7 so DC48. These angle anchors along side directions.

At E ray EA points along side C→A. At F ray FB along A→B. At D ray DC along B→C. These three reference rays are parallel to sides, orientations likely separated 120° cyclic. Lines EP, FP, DP all make common undirected angle with respective reference side direction. Could express P projections / rotation.

Could set three lines through D,E,F whose directions relative to side-reference are common angle. Their pairwise orientation perhaps lines correspond under 120 rotations, and their concurrency P? Need solve concurrency and common angle.

Let's coordinate/vector.

Take A=(0,0), B=(55,0), C=(27.5, 55√3/2) (counterclockwise equilateral). Interior upper-right. F AF40: F=(40,0). E CE30 along C to A: C + (30/55)(A-C) = (12.5, (25)(√3/2)? C y=55√3/2 times25/55=25√3/2), so E=(25/2,25√3/2). D BD7 along B to C: B +(7/55)(C-B): vector (-27.5,55√3/2)*7/55 =(-3.5, 7√3/2), D=(103/2,7√3/2). Great.

Reference:
EA=A-E =(-25/2,-25√3/2), angle 240°.
FB=B-F=(15,0), angle 0°.
DC=C-D: along B→C direction 120°, vector likely (48)(-1/2,√3/2)=(-24,24√3), angle120°.

So reference ray angles β:
at E 240°
at F 0°
at D 120°.

If each angle equal perhaps P lies on a certain side of each ref ray dictated interior. Rays EP, FP, DP likely directions from boundary points inward. Need identify theta and orientation. Since P inside triangle, EP from E into interior relative side CA; FP from F into interior; DP from D into interior. Reference:
- E on CA, EA points along side down-left. Interior is to which side? side C→A, B lies right/east of line, cross? EP likely angle clockwise?
- F on AB horizontal, FB right; interior above, ray FP forms θ above FB => direction θ (0<θ<180 likely)
- D on BC, DC up-left, interior side A is left/down? Reference DC angle120; interior is clockwise side perhaps direction 120-? Let's determine. At BC toward C, A lies left or right? vector DC 120°, A-D (-51.5,-6.06), cross DC direction (-.5,.866) with A-D: (-.5*-6.06) -(.866*-51.5)=3.03+44.6 positive => left side (CCW), so inward ray DP is CCW from DC, orientation 120+θ if θ.
At F inward is CCW from FB, angle θ.
At E: vector EA 240°, B relative? E to B (42.5,-21.65), cross EA unit (-.5,-.866) with B-E: (-.5*-21.65)-(-.866*42.5)=10.8+36.8 positive, inward left (CCW) too? Hmm side E→A vector down-left, B is left side due path C→A maybe triangle CCW? Actually triangle A->B->C ccw. Edge C->A is ccw continuation, interior left. yes. Thus EP orientation 240+θ mod360. For P interior likely this wraps etc. FP θ. DP 120+θ. Since EP base 240 + θ, potentially direction into interior for angles? At E interior wedge from EA angle240 CCW to EC angle60 (through 300,0,60), so θ ∈0..180. FP inward from FB angle0 toward FA180, θ∈0..180. D inward from DC120 to DB300 CCW =180, θ∈0..180. Great. Thus:
direction EP angle φ=240°+θ mod360.
FP φ'=θ.
DP φ''=120°+θ.
These are separated by 120°. Indeed EP direction = FP+240 = FP-120, DP=FP+120. So the three lines from points D,E,F to P have orientations θ + 0, +120, -120. Interesting: A ray from P back to points has opposite orientations.

We need common point P such that vectors P-F, P-E, P-D lie along three rays making 120 directions. Let unit vectors perhaps u0=(cos θ,sin θ); u120=R120 u0; u240=R240 u0.
P=F+t_F u0
P=D+t_D u120
P=E+t_E u240, positive distances.

Unknown θ and lengths. Subtract positions. Geometrically triangle DEF with rays to P in directions rotated common. Could derive P via spiral/similarity/circumcenter? The lines intersect. There may be complex equation and tan² rational perhaps.

We can solve coordinates linear/trig. Use projection equations.

Represent each equality P-F=t0 u. Then P-D=t120 R120u. Difference F-D etc. Dot with normals.

Could exploit sum of unit vectors u+R120u+R240u=0. Maybe distances relation and point weighted sum.

Let x0=PF, x1=PD, x2=PE. Position equations. Since three equal points. Differences side vectors yield:
D-F = x0 u - x1 R120u (because P=F+x0u =D+x1R u)
E? etc. Unknown x and θ. Three lengths, angle, 2D conditions effectively perhaps.

Could use rotations mapping lines and fixed points. If rotate one ray direction to align, transformations center etc. Concurrency condition linear in cos/sin. Solve.

Let c=cos θ,s=sin θ. We know coordinates and reference point differences align with side axes. Could write line equationsusing sine/cos and intercepts. Three lines:
Through F direction angle θ:
-s (x-x_F)+c(y-y_F)=0 => -s x+c y + s*40=0.
Through D direction θ+120:
normal n120=R90 direction? line eq. Through E angle θ+240. Sum perhaps cancellation yields relation involving points, solve s,c.

Each line orientation differs 120. Let define normal n0=(-s,c). Line F: n0·(X-F)=0.
For D direction R120 u, normal R120 n0. line D: (R120 n0)·(X-D)=0.
E: (R240 n0)·(X-E)=0.
Concurrency of three oriented parallel? Their normals sum zero. There is formula constant compatibility.

Could solve first two P as affine in c,s then impose third; maybe equations.

Alternative express line coordinate projections. For each line, q_i=n_i·fixedpoint. Intersection satisfies N P=q with 3x2.

Because n0+n120+n240=0. To be consistent, perhaps q0+q120+q240 plus terms must zero; but n's sum zero and three equations overdetermined: constants must perhaps obey relation. Yet coefficients rows sum zero; if P exists, sum LHS=0, so q constants sum 0 automatically? Let's compute q0+q1+q2 =? each n_i·point. Sum n_i· corresponding different points, not generally zero; it depends θ. This itself gives an equation! Wait if rows exactly sum zero, consistency requires RHS sum zero. Then if this equation holds, first two determine P. Thus concurrency condition simply sum constants=0. Likely yields linear in c,s, enough tan. This is key. For P inside choose θ.

Let's calculate q_F=n0·F; q_D=n120·D; q_E=n240·E. Sum zero.

Coordinates symmetric may simplify.

n0=(-s,c).
n120=R120 n0. R120 matrix [[-1/2,-√3/2],[√3/2,-1/2]]:
n120 = ((s -√3 c)/2, (-√3 s - c)/2).
n240=R240 [[-1/2,√3/2],[-√3/2,-1/2]]:
n240=((s+√3 c)/2, (√3 s-c)/2).

F=(40,0), qF=-40s.
D=(103/2,7√3/2).
qD = ((s-√3c)/2)*(103/2) + ((-√3s-c)/2)*(7√3/2)
=1/4[103(s-√3c)+7√3(-√3s-c)]
=1/4[(103-21)s + (-103√3-7√3)c]
=1/4[82s-110√3 c]= (41/2)s -(55√3/2)c.

E=(25/2,25√3/2).
qE=((s+√3c)/2)*(25/2)+((√3s-c)/2)*(25√3/2)
=25/4[(s+√3c)+(3s-√3c)] =25/4*4s=25s.
Sum q = -40s +(41/2)s -(55√3/2)c +25s = (-40+20.5+25)=5.5s -(55√3/2)c = (11/2)s -(55√3/2)c. Set=0 -> 11s=55√3c => tan θ=5√3, tan²=75. Is it that simple? θ ~83.4°, P perhaps inside. likely answer 75. Need ensure orientation interpretation. Maybe common angles could have differing orientations due P location; but inside dictates as above likely. Need step-by-step verify P exists and lies inside. Concurrency equation guarantees if c positive etc θ acute (tan positive). c could sign; tan. There may be solution θ with tan=5√3 and either θ acute or θ+180, but ray directions θ in (0,180) due interior so θ acute. Need show P inside. Since line consistency yields unique P, should verify.

Let's calculate perhaps. c=1/sqrt(1+75)=1/√76=1/(2√19), s=5√3/(2√19) if θ acute. Find P and barycentric verify.

Line through F:
-s(x-40)+c y=0 => y=(s/c)(x-40)=5√3(x-40).
Line D direction θ+120 etc. Solve perhaps line equation qD. n120·P=qD. Or derive D line. Could calculate P. Need ensure line intersections.

Let k=√3 perhaps. n120=(s-√3c, -√3s-c)/2. Substitute ratio. Compute normalized maybe multiply 2√19:
s=5√3/(2√19), c=1/(2√19).
s-√3c=4√3/(2√19)=2√3/√19.
-√3s-c=-15/(2√19)-1/(2√19)=-8/√19.
n120=(√3/√19,-4/√19). Thus line D: √3 x -4y=qD*√19. qD=(41/2)*(5√3/(2√19)) -(55√3/2)*(1/(2√19)) = ((205-55)√3)/(4√19)=150√3/(4√19)=75√3/(2√19). multiply √19 =75√3/2. So √3 x -4y=75√3/2.

F line y=5√3(x-40). plug: √3 x -20√3(x-40)=75√3/2. Divide √3: x-20x+800=37.5 => -19x=-762.5 => x=762.5/19=1525/38 ≈40.1316. Then y=5√3*(5/38)=25√3/38≈1.139. P near F slightly right, inside above AB. Is it inside relative sides? x~40.13, y 1.139. Left side AB? Equilateral upper, side AC line y=√3 x (A to C); interior below y<=√3x: yes 1.1<69.6. Side BC line? y <=? line B-C y=√3(55-x) for x 40 gives26; yes. So inside. Good. Wait P x>F 40 and line direction θ~83.4 has x increasing slight, yes.

But hold: angle AEP: line EP direction from E to P. Our orientation assumed 240+θ mod360. With θ~83.4 gives323.4°, down-right. E=(12.5,21.65), P=(40.13,1.14): yes down-right angle ~323.4. Angle between EA direction240 and EP323 =83.4. F->P direction83.4, D->P θ+120=203.4 down-left. D=(51.5,6.06) to P=(-11.37,-4.92) direction203.4. Good.

Answer 75. Need provide step-by-step. But let's ensure no subtle angle definition common could have exterior angle or angle >90? Standard ∠ between rays 0-180. P inside imposes each ray in inward wedge and reference side boundary; oriented offsets all same CCW as reason. Thus θ∈(0,180). tan² invariant sign. We can explain.

Need perhaps present an elegant derivation. Need make sure normal labels and calculations correct. Let's thoroughly structure.

Set A=(0,0), B=(55,0), C=(55/2,55√3/2). Then:
F=(40,0) because AF=40.
E: CE=30 means AE=25. Coordinates E = A + (25/55)(C-A) = (25/2,25√3/2). Correct C vector (55/2,55√3/2)*25/55 = (25/2,25√3/2).
D: BD=7. B +(7/55)(C-B). C-B=(-55/2,55√3/2). = (55-7/2, 7√3/2)=(103/2,7√3/2). Good.

Let θ=∠ common. Direction angles:
Need map:
- FB ray from F to B is angle 0.
- DC D→C angle120°.
- EA E→A: C→A direction angle? A-E=(-12.5,-21.65), angle240°.
All interior is to left of these oriented boundary rays? Let's carefully establish each and orientation:
Triangle boundary orientation? A→B→C is counterclockwise. But rays are FB along A→B (part), DC along B→C, EA along C→A. They are indeed forward boundary vectors in CCW polygon A→B→C→A. At any point on side, interior lies to left for CCW traversal. Thus rays to P are obtained by CCW rotations by θ from:
FB angle0 -> FP angle θ.
DC angle120 -> DP angle120+θ.
EA angle240 -> EP angle240+θ (mod360).
Yes. Note the listed angle ∠AEP uses ray EA then EP, exactly CCW; ∠BFP FB→FP; ∠CDP DC→DP. All interior left. θ can range 0 to180 (wedge angles 180 due rays to endpoints? At boundary points, inward open half-plane intersects directions spanning 180; yes).

Define unit normal to direction angle α as n(α)=(-sin α, cos α), i.e rotate +90. Let n0 for θ, n1 θ+120, n2 θ+240. Lines FP, DP, EP:
n0·(X-F)=0
n1·(X-D)=0
n2·(X-E)=0.
Since n0+n1+n2=0. Summing equations:
0 =? If X common, sum n_i·X - sum n_i·points =0, first sum 0, so sum n_i·fixed points=0. Or equations n_i·X=n_i·Q_i; adding gives RHS sum0. Compute.

Could choose notation R perhaps. Need show explicit normals.

n0=(-s,c).

n1=(-sin(θ+120), cos(θ+120)).
Compute:
sin(θ+120)=s cos120+c sin120=-s/2+√3 c/2 => negative = (s-√3c)/2.
cos= c cos120-s sin120=(-c-√3s)/2. correct.

n2=(-sin(θ+240), cos):
sin θ+240=s(-1/2)+c(-√3/2)=-(s+√3c)/2, negative=(s+√3c)/2.
cos=c(-1/2)-s(-√3/2)=(-c+√3s)/2. correct.

Then dot values:
n0·F=-40s.
n1·D:
D x=103/2, y=7√3/2.
First: 103(s-√3c)/4.
second: 7√3(-c-√3s)/4 =(-7√3c-21s)/4.
sum=(82s-110√3c)/4=(41s-55√3c)/2. yes.
n2·E:
first 25(s+√3c)/4
second 25√3(√3s-c)/4=(75s-25√3c)/4
sum=100s/4=25s. yes.

Set:
-40s +(41s-55√3c)/2+25s=0.
Combine s terms (-15s?) -40+25=-15; +20.5 =5.5. Thus 11s/2 -55√3c/2=0 -> 11s=55√3c -> tanθ=5√3, assuming c nonzero. θ likely acute from equation and s>0 implies c>0. θ= arctan. tan²75.

Need verify consistency condition is sufficient; first two lines always intersect except perhaps? θ solution direction nonparallel (0 and120), so unique P. It automatically lies on third due sum. But to establish actual desired P inside, perhaps mention calculate P. The premise guarantees one, but our derivation selects direction based on P inside; no need construct. Yet to reassure and eliminate extraneous from normal sum? We used necessary condition only. There could in principle no θ? But equation gives one and lines intersection; verify inside. Need include perhaps coordinates P to establish. Compute exact P carefully.

At tan=5√3, line through F slope tan θ: y=5√3(x-40), since θ in (0,180) and equation yields c positive because s>0, so acute. Through D direction θ+120 line normal relation. We found √3x-4y=75√3/2. Solve:
y=5√3(x-40)
√3x -20√3(x-40)=75√3/2
x -20x+800=37.5
-19x=-762.5= -1525/2; x=1525/38.
x-40=1525/38-1520/38=5/38.
y=25√3/38.
Check line D with values: √3*1525/38 -100√3/38 =1425√3/38=75√3/2 (1425/19=75) yes.
Inside inequalities:
Triangle with base AB y≥0.
Below AC y≤√3x.
Below BC? line BC y≤√3(55-x) for relevant.
P y positive. Need verify:
√3x - y =√3(1525/38 -25/38)=1500√3/38>0.
√3(55-x)-y =√3(55 -1525/38 -25/38)=√3((2090-1550)/38)=540√3/38>0. So inside. Could simply state these hold, so valid.

But desired response "reason step by step" likely enough. Need put final numerical answer last exact format. Must ensure no text after. Number only after colon. Last line "Final Answer: 75".

Let's see whether any alternate issue: They might expect rational? tan²=75. Let's independently validate via vector distances perhaps. Could there be orientation sign different if P inside but angle at D measured from DC toward DP clockwise rather than CCW? We asserted interior left of CCW boundary sequence. Let's verify DC is B→C, D lies BC; polygon A→B→C is CCW. At edge B→C, interior left indeed. DP from D to P must be left of DC. Thus angle from DC counterclockwise to DP is in (0,180). Standard ∠CDP measures smaller absolute angle, but because P is interior and wedge exactly left 180°, yes equals CCW θ. Similarly E on CA; ray EA is C→A, interior left. Fine.

But check side E and AE=25; coordinate yes. ∠AEP vertex E: ray EA points from E toward A (C→A), forward boundary; yes.

F ∠BFP: ray FB from F to B, forward A→B; yes.

D ∠CDP: ray DC, forward B→C; yes. So all boundary rays consistently oriented around triangle. This clever setup leads normals sum.

Let's consider whether θ could be obtuse. At F ray FB direction0; inward half-plane y>0 means FP direction φ ∈(0,180), so θ=φ, can obtuse. At D line DP orientation θ+120 mod360 must in inward half-plane. For θ in (0,180), θ+120 in (120,300), all are left of orientation120 (relative angle θ), okay. At E θ+240 mod360; for θ<120 gives 240-360, for >120 wraps 0-60, inward relative side yes. Thus all. Equation tan positive; with s positive, c positive so acute. Good.

Need make explanation of normals and row sum clear. Let's perhaps derive rotations without overly complex.

Potential issue: Normal vectors n_i sum zero? n(θ)+n(θ+120)+n(θ+240)=0 indeed for any θ. We can state rotating by 120° thrice.

Let's write final in coherent steps:

1 coordinates.
2 orient rays.
3 line equations and concurrency.
4 solve and optionally verify P inside.
5 answer.

Let's ensure final numerical answer line exact format. User says number only no units, likely last line 'Final Answer: 75'. Before that explanatory text okay.

Let's produce with LaTex.

Detailed coordinate calculation:
A=(0,0), B=(55,0), C=(55/2,55√3/2).
D:
D=(55,0)+(7/55)(-55/2,55√3/2)=(103/2,7√3/2).
E=(25/55)C=(25/2,25√3/2). Since AE=25.
F=(40,0).

Direction angle table perhaps:
ray | boundary direction | ray to P:
FB 0, FP θ
DC 120°, DP θ+120°
EA 240°, EP θ+240° mod360.
Explain CCW interior.

Normals:
m maybe n0,n1,n2. Then line equations:
n0·(X-F)=0 etc.
Consequence:
n0·F+n1·D+n2·E=0.
Wait sign: line n_i·(X-Q_i)=0. Sum: (sum n_i)·X -sum n_i·Q_i=0 => sum n_i·Q_i=0. correct.

Then expressions and simplify.

Check P verification coordinates perhaps line through D equation derived; include only concise. Let's recompute c,s normalization:
tan=5√3. Since s>0 and relation 11s=55√3c implies c>0. So θ acute. Set c=1/√76, s=5√3/√76 = 5√3/(2√19). Earlier yes because √76=2√19. Direction n120 etc. Line D equation:
n1·X=n1·D.
n1 components after scaling:
s-√3c = (5√3-√3)/(2√19)=4√3/(2√19)=2√3/√19. Dividing by 2 -> √3/√19.
(-√3s-c)/2: -(15+1)/(2√19)/2? √3s=15/(2√19), c=1/(2√19), sum16/(2√19)=8/√19; /2 =4/√19 negative. So n1=(√3,-4)/√19.
RHS n1D=75√3/(2√19). So line √3x-4y=75√3/2. Correct.
F line n0·(X-F)=0: -s(x-40)+cy=0 -> y=(s/c)(x-40)=5√3(x-40).
Solve x=1525/38. Verify:
√3 x -4*5√3(x-40)=√3[x-20x+800]=√3(800-19x)=75√3/2 => 800-19x=37.5 =>19x=762.5=1525/2 => x=1525/38. y=5√3*5/38=25√3/38.
Inside:
Triangle is set y≥0, y≤√3x, y≤√3(55-x).
At P:
√3x-y = √3(1525-25)/38=1500√3/38 >0.
√3(55-x)-y = √3[(2090-1525-25)/38]=540√3/38 >0. Good.

Actually if include this, demonstrate no extraneous. But perhaps no need; premise says exists. Yet good rigor.

Let's think if line consistency equation could also be satisfied if c negative θ quadrant II? θ by definition direction from FB to FP in (0,180), s>0. Equation forces c>0, no alternate. Fine.

One subtlety: Standard angle AEP might be >? θ between EA and EP; if EP direction relative EA can be 180? no interior. If P exactly side? inside not boundary, θ ∈(0,180). Good.

Let's consider alternate coordinate orientation might yield equation tan=- etc but tan² same maybe. No issue.

We need not mention all exact normal trig perhaps show.

Let's prepare final answer with headings maybe.

But the requested desired oververbosity 5, step by step. We'll deliver around 700-1000 words? Likely 600. Let's ensure no unsupported leap.

Let's articulate why interior to left: coordinate triangle has A→B→C counterclockwise. Along each of oriented rays FB, DC, EA, we're traversing sides in that same CCW order. Therefore P lies to left, and rotating each reference ray CCW by θ yields ray toward P. Note ∠ values equal θ. This is central.

Then table.

Let's make normal convention n(α)=(-sinα, cosα). It is perpendicular to unit ray direction (cosα,sinα). For lines through fixed points:
n0 normal FP, n1 normal DP, n2 normal EP. Sum equations. Since normals sum zero.

Compute:
n1 D:
\[
\frac{s-\sqrt3c}{2}\frac{103}{2}+\frac{-\sqrt3s-c}{2}\frac{7\sqrt3}{2}
=\frac{41s-55\sqrt3c}{2}.
\]
n2 E as above.
Then sum.

Check n0 F uses F=(40,0), n0=(-s,c), -40s.

Sum simplification:
\[
-40s+\frac{41s-55√3c}{2}+25s
=\frac{11s-55√3c}{2}.
\]
Yes because -40+25=-15, -30/2 +41/2=11/2. Good.
Thus.

At verification, maybe not necessary to explicitly write all inequalities but nice. If no verification, one could question that equation is only necessary. But problem gives existence and our orientation makes θ unique; yet final should establish. We'll include coordinate P and inequalities compactly.

Let's ensure line D equation is valid and not accidentally line to P orientation could be opposite due normal no orientation; the line equation allows opposite ray too. Intersection P we calculate indeed matches correct rays, verified inside. Good.

Let's check P relative ray from F: P-F=(5/38,25√3/38), angle tan = (25√3)/(5)=5√3, positive direction. Correct.

Potential alternate line intersection may have P but if line equations no direction, and sum condition could produce other intersections? Only unique P. Verify is inside, so correct ray. Fine.

Let's perhaps state first two lines intersect because directions differ 120°, so no parallel issue. The sum condition ensures third. This makes sufficiency explicit.

Could skip exact c,s and derive D line more directly using angle line slope. But verification as above requires trig. Fine.

Let's draft final:

Let θ common.

Coordinates...

Normals and equations...

After tan result:
"For completeness, this angle indeed gives an interior point..." Then derive. Last line.

Need put final numerical answer on last line in exact format. Avoid another math after. Good.

Let's produce.


Question: Let $b \geq 2$ be an integer. Call a positive integer $n$ $b\textit{-eautiful}$ if it has exactly two digits when expressed in base $b$ , and these two digits sum to $\sqrt{n}$ . For example, $81$ is $13$ -eautiful because $81=\underline{6}$ $\underline{3}_{13}$ and $6+3=\sqrt{81}$ . Find the least integer $b\geq 2$ for which there are more than ten $b$ -eautiful integers.

Let \(n\) be \(b\)-eautiful, and write its two base-\(b\) digits as

\[
n=ub+v,
\]

where

\[
1\le u\le b-1,\qquad 0\le v\le b-1.
\]

Let

\[
s=u+v=\sqrt n.
\]

Then \(n=s^2\), so

\[
s^2=ub+v=ub+(s-u)=u(b-1)+s.
\]

Therefore

\[
s(s-1)=u(b-1).
\]

Set \(t=b-1\). Thus \(s\) must satisfy

\[
s(s-1)\equiv 0\pmod t.
\]

Also, since \(n=s^2\) has exactly two digits in base \(b=t+1\),

\[
t+1\le s^2<(t+1)^2,
\]

so

\[
\left\lceil\sqrt{t+1}\right\rceil\le s\le t.
\]

Conversely, any \(s\) in this interval satisfying \(t\mid s(s-1)\) determines valid digits \(u\) and \(v\).

Now count the solutions modulo \(t\). If

\[
t=\prod_{i=1}^{\omega(t)}p_i^{a_i},
\]

then for each prime power \(p_i^{a_i}\), because \(s\) and \(s-1\) are relatively prime, the entire prime power must divide either \(s\) or \(s-1\). Thus, for each distinct prime factor, there are two possibilities. By the Chinese remainder theorem, the total number of solutions modulo \(t\) is

\[
2^{\omega(t)}.
\]

Among these solutions, \(s=1\) is below the required interval. Every other solution satisfies \(s\ge \lceil\sqrt{t+1}\rceil\): indeed, if \(2\le s<\sqrt{t+1}\), then

\[
0<s(s-1)<t,
\]

so \(s(s-1)\) cannot be divisible by \(t\). Hence the number of \(b\)-eautiful integers is

\[
2^{\omega(b-1)}-1.
\]

We need this number to exceed \(10\). If \(\omega(b-1)=3\), the number is \(2^3-1=7\), while if \(\omega(b-1)=4\), it is \(2^4-1=15\).

Therefore \(b-1\) must have at least four distinct prime factors. The smallest such integer is

\[
2\cdot3\cdot5\cdot7=210.
\]

Thus the least possible base is

\[
b=210+1=211.
\]

Final Answer: 211

Explanation:
We need solve, step by step. Need carefully derive. Need least b such that count >10 b-eautiful integers.

Let n two digits base b: n= x b + y, x in {1,...,b-1}, y {0,...,b-1}. Digits sum s=x+y = sqrt(n), so n=s^2. Conditions x+y=s, s^2 = xb+y = xb+(s-x)=x(b-1)+s. Thus s^2-s=x(b-1), so s(s-1)=x(b-1), with x integer digit, and y=s-x in [0,b-1], x in [1,b-1]. Also n=s^2 exactly two digit implies b <= s^2 <= b^2-1 perhaps x positive assures >=b and b<=b^2 etc. Need count valid s maybe.

Equation means d? Let k=b-1. x=s(s-1)/k. Require 1<=x<=k=b-1 and y=s-x 0..k. Since s perhaps ≤2k because digit sums max 2k, also x positive. Conditions y≥0 gives s≥x. y≤k gives s-x≤k. x≤k. Equation.

Set x=a. Then b= a? s roots. s(s-1)=a(b-1). For integer s satisfying quadratic. b=1+s(s-1)/a. Count per b.

Could parameterize divisors. Need threshold counting.

Alternative let quotient q=s-x = y. Equation? s=x+y and (x+y)^2=xb+y => x(x+2y)+y^2=xb+y => x(x+2y-b)=y-y^2? Better use s.

Set t = b-1. Need s(s-1) divisible by t, quotient x, and inequalities.

From x=s(s-1)/t. Since s? n=s² two-digit in base b=t+1 gives t+1 ≤s² <(t+1)^2. Lower s≥ceil sqrt(t+1), upper s≤t (since strict < (t+1)^2 means integer s≤t). Is upper automatically digit sum max 2t, but n sqrt < t+1 => s≤t. This simplifies y=s-x and x≤t.

Lower bound s²≥t+1. y=s-x in range perhaps if 0≤s≤t and x?Equation x=s(s-1)/t. For s≥1. x≤? s(s-1)/t ≤ t for s≤t+? s≤t gives max t-1, yes. Positive for s≥2. y=s-x = s -s(s-1)/t = s(t+1-s)/t = sb/t. Need y≤t: s≤t²? yes for s≤t; y≥0. So for s in [ceil sqrt(t+1), t], divisibility t | s(s-1) yields unique valid pair and number. Is there concern y integer yes quotient and expression; x integer, y=s-x. And x≥1 for s≥2. Great. Thus count C(b)=# integers s with ceil sqrt b ≤ s≤b-1 and (b-1)|s(s-1).

Let t=b-1. Lower ceil sqrt(t+1). Let r = ceil sqrt(t+1). Need # s ∈ [r,t] with s(s-1) ≡0 mod t.

Could shift residues via s perhaps t divisor product consecutive. Let a=s? Count modulo t. s ranges t-r+1 numbers. Trivially s=1,t are divisibility roots, but only t likely in range unless r=1 no. So always at least one (s=t), yielding n=t²? b= t+1, representation? t²=(t-1)(t+1)+1 digits t-1,1 sum t yes. Example 81 b13 t12 s9, not universal. Universal s=t=12: n144 representation 11,1 sum12 yes.

Need least b where >10 solutions. We need characterize and search potentially huge. Count roots s modulo t in interval length t-r+1 nearly t, with excluded residues 1,...,r-1 approximately sqrt b. Total roots congruences count depends on factorization. For modulus t, number of residues s mod t satisfying s(s-1)=0 is 2^{ω(t)} if t is squarefree? More generally CRT each prime power p^a must divide either s or s-1 due consecutive gcd 1, so exactly 2 choices per distinct prime factor; roots count modulo t =2^{ω(t)}. Indeed for every prime power allocated entirely to one factor, CRT 2 choices per distinct prime factor; count exactly 2^ω. Residues 0 and1 etc. Thus full cycle [1,t] has R=2^{ω(t)} roots. We count range [r,t], equivalent exclude s=1,...,r-1. Need at least 11 roots in latter. If total R>11 automatically, since exclusion can at most? Full roots >11 means ≥12 powers of 2, i.e. ω≥4 ->16, yes all range has at least R-(r-1), but could exclusions huge r; not automatic. Wait roots excluded among small numbers. There are only residues small integers satisfying divisibility t | s(s-1). Since s<r≈sqrt t, product s(s-1)<t likely except boundary, so only s=1 is divisible (product 0). Let's check if s≤r-1 < sqrt(t+1), then for s≥2, 0<s(s-1)< (sqrt(t+1)-1)(sqrt...-2)<t, cannot divisible. So exactly s=1 excluded. Great! Thus count C=2^{ω(t)}-1 (root s=1 excluded), for t? All root residues distinct 1..t. Since range excludes 1 only among roots. Need verify r could 1 t=0 no. b≥2 t≥1. For t=1, roots mod1 count 1, formula 2^0=1, range [ceil sqrt2=2? t=1 empty], C formula 0 works. For t=2: roots s mod2 all? s(s-1) even every s, 2 roots, exclude 1, range ceil sqrt3=2 to2 gives s2 yes C1. General yes s among 2..r-1 product less t? r-1 < sqrt(t+1); (r-1)(r-2) < t maybe for t=2 r=2 none. Fine.

Thus need least b such that 2^{ω(b-1)}-1 >10. Powers: ω=3 gives 8-1=7; ω=4 gives16-1=15. Need b-1 have at least 4 distinct prime factors. Least integer t with ω(t)≥4 = product first four primes 2*3*5*7=210. Then b=211. Is it that simple? Check condition >10 count 15. Any t <210 could have 4 distinct prime factors? Minimal is 210, no. So answer 211. Need reason step-by-step and perhaps explicitly verify 15 integers maybe not list neededbut ensure subtle issue: root count modulo t exactly 2^{number distinct primes}, including non-squarefree. CRT: For t=∏p_i^{a_i}, congruence s(s-1) divisible by p_i^{a_i}; because s,s-1 coprime, all p_i^a divides exactly one. For each i, 2 choices modulo p_i^a. CRT gives 2^ω residues mod t. Yes.

Our range roots and exclusion proof. Let t=b-1, n=s². Derive digit relation. Let digits A,B maybe avoid b confusion. n has exactly two base-b digits means n=ub+v with 1≤u≤t,0≤v≤t. Sum u+v=s and s²=n. Then:
s²=ub+v=ub+(s-u)=u(b-1)+s=ut+s, so u=s(s-1)/t.
Conversely, if s in something and divisibility, define u quotient and v=s-u; need prove digits within bounds and n exactly two digits. We set s from ceil sqrt b through t. Need ensure v ∈[0,t]. Derive v=s-u=s -s(s-1)/t=s(t+1-s)/t =s(b-s)/t. For s in [?,t], nonnegative, and ≤t? s(b-s) ≤? with b=t+1. For s between 1,t, product max maybe floor((t+1)^2/4), can exceed t for t>3! Wait v=s(t+1-s)/t. Need check ≤t condition: s(t+1-s) ≤ t². Is this always for 1≤s≤t? Let f=s(t+1-s). For s around (t+1)/2, f≈(t+1)^2/4, which for t>3 is >t²? Compare (t+1)^2/4 ≤t² iff t+1≤2t iff t≥1 (equality t=1), yes! For t≥1, max (t+1)^2/4 ≤ t² for t≥1 indeed because t²*4 - (t+1)^2 =3t²-2t-1=(3t+1)(t-1), ≥0. So yes v≤t. Integrality obvious. u≤t? s(s-1)/t for s≤t: max t-1, yes. u≥1 for s≥2. Thus valid.

But perhaps lower bound should be ceil sqrt b, and b=t+1. Since n=s² exactly two base b digits: s²≥b and s²<b². s≤b-1=t. Conversely if s in that interval, s² in [b,t²], and t² < b² = (t+1)², so exactly two. Yes.

Then roots in [r,t]. Full residues 1..t. Need show only root excluded is s=1. For s in [2,r-1], s²<b by r definition? r=ceil sqrt b. If s≤r-1 then s<sqrt b, so s²<b=t+1, hence s²≤t. But product s(s-1)<s²≤t, and positive, so cannot divisible by t. More directly. So only s=1 root below r. Thus C= roots total -1.

Check edge t=1: full residues one s=1; lower r=ceil sqrt2=2 >t, count 0. Formula total 1 minus1=0. Good.
t=2 roots modulo2: s all, root count 2 formula 2^1=2; exclude 1 count1.
At b=211, t=210=2*3*5*7 squarefree ω4, roots total16 modulo210. Exclude 1 gives15. All remaining roots in [ceil sqrt211=15? sqrt211≈14.526 ceil15, ...,210]. Need ensure none roots 2..14: product less210 except s=15 product210 divisible and r=15. Yes 15 included. Count 15 >10.

But wait perhaps count of roots modulo t is 2^{ω(t)} only if prime powers and each p_i^a divides one factor. Correct. But for each i exactly two congruence classes modulo p_i^a: s≡0 or1. CRT. If t maybe no? yes.

Then minimize b with count>10. Count values: If ω≤3, C≤7. If ω≥4, C≥15. First t with 4 distinct primes is 210. Therefore b=211. Is it possible t has ω>4 less? no, minimum first four. Good.

Need present step-by-step. User explicitly asks reason step by step and final line exact format "Final Answer: <number>" number only no units. They mean final numerical answer line format.