Russian Math Olympiad | A Very Nice Geometry Problem | Find the area of the square

[@ناصريناصر-س4ب]

QD²=QP*QB and from it QD²=2*6=12 so QD=√12 and according to the Pythagorean theorem (a-√12)²+a²=6² so a=√3+√15 and from it the area of the square (√3+√15)²=18+6√5

[@fouadyoussef3113]

This is the way I did.

[@ناصريناصر-س4ب]

Yes, as soon as you see a circle, a secant, and a tangent, this theory comes to your mind, although there are many ways to solve it.​@@fouadyoussef3113

[@RAG981]

I agree, much more straightforward.

[@ناصريناصر-س4ب]

Yes, as soon as you see a circle, a secant, and a tangent, this theory comes to your mind, although there are many ways to solve it.

[@fouadyoussef3113]

@ناصريناصر-س4ب اخ ناصر حضرتك من اين؟

[@marcgriselhubert3915]

Let's use an orthonormal center B and first axis (BA), let's name c the side length of the square and a = CQ. Equation of the circle: (x - c)^2 + y^2 = 0 or x^2 + y^2 -2.c.x = 0, equation of (BQ): y = (c/a).x At the intersection (point P) we have x^2 +(c^2/a^2).x^2 -2.c.x = 0 or ((a^2 + c^2)/(a^2)).x^2 - 2.c.x = 0. As x0 at point P and as a^2 + c^2 = 36, we have: x = (c.(a^2))/18 and then y = (c/a).x = (a.(c^2))/18 VectorBP = ((a.c)/18).VectorU with VectorU(a, c), then BP = ((a.c)/18).sqrt(a^2 + c^2) = ((a.c)/18).6 = (a.c)/3 Knowing that BP = 4, we then have that (a.c)/3 = 4 and then a = 12/c. As a^2 + c^2 = 36, we theh have (144/(c^2)) + (c^2) = 36 or c^4 -36.(c^2) +144 = 0. Deltaprime ) 18^2 -144 = 180 = (6.sqrt(5))^2 Finally c^2 = 18 -6.sqrt(5) (rejected *) or c^2 = 18 + 6.sqrt(5) which is the area of the square. (* if c^2 was 18 - 6.sqrt(5) which is about 4.58, then c would be about 2.14 and the diagonal BCof the square would be about 3.02 which is inferior to BQ = 6, so it is impossible. With c^2 = 18 + 6.sqrt(5) BC is about 7.92 which is OK.)

[@Z-eng0]

Could've been done much faster using the tangent-secant theorem of circles that'd go: QD² = QP * QB. Then going for a right triangle with QC = DC - QD. And use Pythagoras theorem to get BC

[@五十嵐特許事務所]

In ⊿CBQ, X = BQcosθ = 6cosθ...(1). Also, in ⊿BAM, ABsinθ = Xsinθ = BM = 2. ∴Xsinθ = 2...(2). From (1) and (2), 6cosθsinθ = 2. ∴3sin2θ = 2. ∴sin2θ = 2/3. This gives cosθ = (√15 + √3)/6, so from (1), X = √15 + √3. The area of ​​square ABCD is X^2, so the desired area is X^2 = 18 + 6√5.

[@alainpeugny1146]

sin θ = 2/x and cos θ = x/6 Thus, 4/x2 + x2/36 = 1

[@hadigayar6786]

DQ^2=QP.QB DQ=root 12 (x-root12)^2+x^2=6^2 x=root3+root15(solved)

[@santiagoarosam430]

Diámetro horizontal =r+r=EA+AB---> Razón de semejanza entre los triángulos QCB y BPE: s=6/2r=3/r---> QC=s*BP =12/r---> (12/r)²+r²=6²---> r²=18+6√5 = Área del cuadrado ABCD. Gracias y saludos.

[@kateknowles8055]

This problem is a big teaser. It is another of these problems where many constructions are possible and few are useful; Or many unknowns might have temporary labels and give a tremulous security. Labelling the midpoint of PB as M and extending AM to R on the quarter circle and S on BC and labelling AB as radius r and labelling MR as m PB.PB= (r+r-m)(m) = 2.2 =4 2rm = 4+m.m r = (4+m.m)/(2m) sin(PAM)= 2/r = 4 m/(4+m.m) cos (2x) =1- 2 (sin(x))(sin(x) cos(PAB)= 1 - 8/(r.r) Using cosine rule in triangle PAB r.r +r.r - 4.4 = 2. r.r. (1 - 8/r.r) 2r.r cancels, changing sides (=changing signs) 16 = 16 which is known already A tangent might be drawn through R. It will be parallel to QB at a distance of m. These are just a few of the places I make to trip myself up. But if I watch the video and read the comments then in a corner of cyberspace I find good (international) company.

[@זאבגלברד]

There is the theorm that since QD is tangent, then QD^2 = 2*(2+4) so QC is a - 12^0.5 so pitagoras in BCQ ....

[@Rudepropre]

Other methods are tangent secant or complete the circle

[@sarantis40kalaitzis48]

Take a look Real.... method. Every comment of him is a new theorem. ..😂😂

[@ВерцинГеториг-ч5ь]

Carry oun AQ , AP=X and perhendicular QK=X to AB , from triangle ABQ by Stewarts theorem - AP*2=(BP/BP+PQ)AQ*2+(PQ/BP+PQ)AB-BPxPQ , X*2=(4/6)AQ*2+(2/6)X*2-4x2 (1) , AQ*2=(AB-\|(BQ*2-QK*2)*2+QK*2=(X*2+6*2-X*2-2X\|(6*2-X*2)+X*2=X*2+36-2X\|(36-X*2) , substitutsng into eguation (1) - X*2=(2/3)(X*2+36-2X\|(36-X*2)+(1/3)X*2-8 , 3X*2=2X*2+72-4X\|(36-X*2)+X*2-24 , 4X\|(36-X*2)=48 , (X\|(36-X*2))*2=(12)*2 , X*4-36X*2+144=0 , (X*2)12=18+-\|18*2-144=18+-\|180=18+-6\|5 , X*2=18+6\|5 , the second root does not satisfy the conditions of the problem since X is less than 4 .

[@dickroadnight]

Yes, I used the cos (supplement) rule (basis of stewart’s Theorem) in AQB.

[@oscarcastaneda5310]

I named DQ as x and found x= 2sqrt(3). Then (r - x)^2 + r^2 = 36 which leads to r^2 = 18 + 6sqrt(5).

[@RealQinnMalloryu4]

(2)^2(4)^2={4+16}=20 {90°A90°B+90°C+90°D}=360°ABCD/20=180ABCD 2^90 2^3^30 2^3^3^10 2^1^3^5^5 2^1^3^2^3^2^3 1^1^1^1^1^2^3 2^3 (ABCD ➖ 3ABCD+2).

[@giuseppemalaguti435]

AQ^2=r^2+(r-√(36-r^2))^2...AQ^2=r^2+4-4rcos(180-arccos(2/r))..(teo.dep coseno)..quindi dopo i calcoli risulta r^2=18+6√5..sempre se i risultati sono corretti

[@فاطمه0محمد-ب8ر]

في الله والله ما كتبت هذا الكلام الا من الضيق وقسوت الضروف اني وامي واخوني نموت من الجوع اني طالبه من الله ثم منك لاتردني خايبه حسبنا الله ونعم الوكيل في من اوصلنا الا هاذا الحال يِآ نآس يِآآمٌـٍة مٌحًمٌد صِآرتٍ قلّوبگٍمٌ بلّآ رحًمٌهً ولّآشفُقهً ولّآ آنسآنيِهً گٍمٌ شگٍيِتٍ وگٍمٌ بگٍيِتٍ گٍمٌ نآديِتٍ وگٍمٌ نآشدتٍ ولّگٍن لّآ حًيِآٍة لّمٌن تٍنآديِ هًلّ يِرضيِگٍمٌ آن آخوآنيِ يِبگٍون ويِمٌوتٍون مٌن آلّجُوع وآنتٍمٌ مٌوجُودون يِعلّمٌ آلّلّهً آلّعلّيِ آلّعظَيِمٌ آننآ لّآ نمٌلّگٍ حًتٍى قيِمٌـٍة گٍيِلّو دقيِق آبيِ مٌتٍوفُيِ اخي اول كلامي انا اقسم بالله على كتاب الله اني لااكذب عليك ولا انصب ولا احتال اني بنت يمنيه من اليمن نازحين من انا واسرتي بيننا ایت الشهرب 20 الف يمني والان علينا 60 الف حق 3 شهور وصاحب البيت من الناس الي ماترحم والله يا اخي انه يجي كل يوم يبهدلنا ويتكلم علينا ويريد من البيت للشارع لانناماقدرنا ندفعله الأجار شافونا الجيران نبكي ورجعو تكلمو الجيران ومهلنالاخره الأسبوع معادفعنا له حلف يمين بالله هذا بيخرجنا إلى الشارع رحمه واحنا. بلادنا بسبب هذا الحرب ولانجد قوت يومنا وعايشين اناوامي واخوتي سفار والدنا متوفي الله يرحمه ومامعنا أحد في هذا الدنيا جاانبنا في هذه الظروف القاسيه اخوتي الصغار خرجو للشارع وشافو الجيران ياكلو واوقفو عند بابهم لجل يعطوهم ولو كسره خبز والله الذي له ملك السموات والارض انهم غلفو الباب وطردوهم ورجعو یبکو ایموتو من الجوع ما احد رحمهم وعطلة ردها لقمت عیش والان لوما احدنا ساعدنا في إيكيلو دقيق اقسم بالله انموت من الجوع فيا اخي انا دخيله على الله ثم عليك واريد منك المساعده لوجه الله انشدك بالله تحب الخير واتساعدني ولو ب 500 ريال يمني مع تراسلي واتساب على هذا الرقم 00967711528949 وتطلب اسم بطاقتي وترسلي ولاتتاخر وايعوضك الله بكل خير اخواني سغار شوف كيف حالتهم وساعدنا وأنقذنا قبل أن يطردونا في الشارع تتبهدل أو نموت من الجوع وانا واسرتي نسالك بالله لولك مقدره على مساعد لاتتاخر علينا وجزاك الله خيرا.~π~π÷~π~÷~π~÷~π~π¢€€°€€°^^^°√•