This Russian Math Olympiad Problem Will Blow Your Mind

Russian Math Olympiad Problem | A Very Nice Geometry Challenge

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[Irtsak]

*The beauty of Geometry*
DZ⊥AB, AP⊥BC, EH⊥AC (Construction)
*When the angle of a right triangle is equal to 30°, remember that the length of opposite side is always equal to half of the length of the hypotenuse.*
So DZ=5/2, EH=2.
*Math Booster* estimated ∠BAC=120° and ∠ BAP =60° (AP bisector ∠BAC)
∠BAP= ∠DAE=60° => ∠BAD+ ∠DAP= ∠DAP+ ∠PAE => ∠BAD= ∠PAE
Working accordingly ∠ΕAC= ∠DAP.
Right triangles AZD, APE are similar => AD/AE=ZD/PE (1)
Also right triangles ADP, AEH are similar => AD/AE=DP/EH (2)
(1), (2) => ZD/PE=DP/EH => ZD⋅EH=PE⋅DP => 5/2⋅2=PE⋅DP => PE⋅DP=5 =>
*PE=5/DP (3)*
BP=PC => 5+DP=PE+4 => *PE-DP=1 (4)*
Solve the system of equation (3), (4) and you will find :
DP=(√21-1)/2 and PE=(√21+1)/2
At last x=DP+PE= …..= √21

*Αεί ο θεός γεωμετρεί (Πλάτων)*
Plutarch attributed the belief to Plato, writing that "Plato said god geometrizes continually" . In modern times, the mathematician Carl Friedrich Gauss adapted this quote, saying "God arithmetizes".

[Steven-vl]

minor issue:: technically, you can't actually "construct" (using a straight-edge and compass) angle α. But you can use a compass to sweep length AD from point A, and length BD from point C; and mark the intersection point as F. Triangles ABD and ACF are congruent by side-side-side, so you can transfer angles α and 30° to triangle ACF.

[prime]

There is some methodolgy.Did you drop an altitude and fail? What else did you try before succeeding. That is a teaching moment!!

[timc]

I wish I had the tutor's imagination! Still, I think I have another (less elegant) solution.
let 't = angle BAD', (the tutor's alpha) and let 'AM = h' be perpendicular to BC( = 9 + X)
tan(30) = 2h/(9 +X) = 1/[sqrt(3), EQ. 1, then
2h = (9 + X)/[sqrt(3)], Eq, 2
tan(ADM) = tan(t + 30) = 2h/(X - 1), Eq.3,
Angle(XAC) = 120 - (60 + t), = 60 - t, so
tan(MXD) = tan(90 - t) = 2h/(X + 1), Eq.4
From EQ. 1, 2, 3, and tan formula: tan(t) = [tan(t + 30) - tan(30)]/[1 + tan(t + 30)tan(30)], and after simplifying:
tan(t) = 5sqrt(3)/[3 +2x], Eq.5, and
From Eq.1, 4: tan(90 - t) = [9 + X]/[{sqrt(3)}(X + 1)], and since tan(90 - t) = 1/tan(t):
[9 + X]/[{sqrt(3)}(X + 1)] = [3 + 2X]/[5sqrt(3)], giving
5(9 + X) = (3 + 2X)(X + 1), or
45 + 5X = 3 + 5X + 2X^2, so
X^2 = 21

[prime]

All you did was rotate triangle ADB externally so that AB coincides with AC.The question is why? What is the plan?

[francescoraucea]

Soluzione così tanto complessa da portare a risultato scorretto. Chiamati alfa e beta rispettivamente gli angoli BAD ed EAC si ottiene alfa+beta=60°; bisecando l'angolo BAD si ottiene una sicura perpendicolare al lato BC che divide la figura in due triangoli perfettamente uguali e le cui basi hanno entrambe lunghezza (5+4+X)/2; la prima soluzione di X che consente la divisione è 3 con quindi lato CB = 12 e lati AB=AC=6, 93 e bisettrice di BAD =3, 47 ; si verifica che tali dimensioni verificano il teorema dei seni.Tanto dovevo al buon senso.

[giuseppemalaguti]

x^2=21...sempre col teorema dei seni, applicato più volte

[ducduypham]

Construct O in the interior of angle ADE such that angle DAO=alpha and AO=AB=AC, then we can prove angle OAE=beta. We can easily prove that triangle ABD congruent to triangle AOD so DO=DB=5 and angle AOD=angle ABD=30°. Also triangle AEO congruent to triangle AEC so EO=EC=4 and angle AOE=angle ACE=30°. We have angle DOE=angle DOA+angle AOE=30°+30°=60°. Using cosine rule in triangle ODC with OD=5, OC=4, angle DOE=60° we can find x=DE=sqrt(21)

[murdock]

This looks amazing - but in my opionion it is not fully correct. Reason: EP ≠ x = DE
φ = 30°; CAB = 4φ; BCA = ABC = φ; BC = 9 + √21 = 2r
BD = 5; CE = 4 → AB = AC = r√3
cos⁡(4φ) = -cos⁡(2φ) = -sin⁡(φ) = -1/2 → (BC)^2 = 6r^2(1 - cos⁡(4φ)) →
(BC)^2/6r^2 = 1 - cos⁡(4φ) = 3/2 ≠ 6(17 + 3√21)/9(17 + 3√21) = 2/3
In your example:
sin⁡(α) = (√2/8)√(17 - 3√21) → α ≈ 18, 59°
sin⁡(β) = (√7/14)√(14 - √21) → β ≈ 35, 47° → α + β ≠ 2φ
It is obvious that ∆ ECP ≠ ∆ ADE and AD ≠ 5 as well as AE ≠ 4.
φ = 30°; AD = a; AE = b; DE = x; CE = 4; CP = 5; ECP = 2φ
DAB = α; CAE = β → α + β = 2φ; DE = x
with law of cosines and law of sines you'll get
α = 36, 74°; β = 23, 26° → α + β = 2φ → a = 4, 18; b = 5, 1 →
a/b = 0, 819 ≠ 4/5 → x = 4, 71 → EP = y = √21 = 4, 58 ≠ x = 4, 71
Please do us a favor and check again, Sir.

[michaeldoerr]

I am now inspired to make use of Math Booster's imagination as a way of practice. I think that the way Geometry has been taught has precluded that ability. And I prefer the solution shown. Some memorization with spatial reasoning. I guess that is the trick that Math Booster makes use of.