Math Olympiad | Can you find Perimeter of the Green triangle ABC? |

[@kaziummeruman524]

Like for preMath👍

[@PreMath]

Thanks dear ❤️

[@kaziummeruman524]

@@PreMath I appreciate it🥰

[@res5139]

Love the step by step detailed solution!! More of such problems please!!

[@jamestalbott4499]

Thank you for the solution, adding the additional lines identifying the congruent triangles! Appreciated this experience!

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@jimlocke9320]

Solution by tangent double angle formula tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)). Let

[@PreMath]

Excellent! Thanks ❤️

[@geraldgiannotti8364]

Excellent! Tangent summation formula. Best solution I have seen.

[@marcelowanderleycorreia8876]

I did like this too. 👋

[@marcelowanderleycorreia8876]

Great question professor!! 👍

[@PreMath]

Glad you liked it! Thanks ❤️

[@jamesthiel4612]

Thanks for the videos. You are very helpful!

[@PreMath]

Glad you like them! You are very welcome! Thanks ❤️

[@LuisdeBritoCamacho]

I solved this tricky Problem using the "Law of Cosines" in triangle [ACD], in the following manner: 1) Let BD = x ; CD = a ; AC = d 2) a^2 = 36 + x^2 3) d^2 = 36 + (10 + x)^2 4) 10^2 = a^2 + d^2 - (2 * a * d * cos(45)) 5) 100 = a^2 + d^2 - (2 * a * d * sqrt(2)/2) 6) 100 = 36 + x^2 + 36 + (10 + x)^2 - (a * d * sqrt(2)) 7) 100 = 72 + x^2 + 100 + 20x + x^2 - (2 * a * d * sqrt(2)/2) Now: if a = sqrt(36 + x^2) ; and d = sqrt(36 + (10 + x)^2) 8) 0 = 72 + 20x + 2x^2 - [sqrt(2) * sqrt(36 + x^2) * sqrt(36 + (10 +x^2))] 9) Two Solutions : x = - 12 and x = 2 So: BC = 6 and BD = 2 so we can conclude that AB = 12 AC^2 = 144 + 36 ; AC^2 = 180 ; AC = sqrt(180) ; AC = 6*sqrt(5) Perimeter = 12 + 6 + 6*sqrt(5) sq un = 18 + 6*sqrt(5) ~ 31,416 sq un Answer: Perimeter is equal to (18+6sqrt(5)) Square Units or Perimeter is approx. equal to 31,416 Square Units.

[@Copernicusfreud]

That is how I did it too.

[@PreMath]

Thanks ❤️

[@alexniklas8777]

From D we draw a perpendicular to AC, point E. By similarity ∆ABC~∆ADE we determine: AB=12; AC=6√5. Perimeter P=18+6√5 Thanks sir!

[@PreMath]

Very good! You are very welcome! Thanks ❤️

[@mariopopesco]

My method : Angle DCB = a In triangle BCD : CD = 6 /cos a Angle BAC is 90 - 45 - a Law of sines in triangle ACD : CD / sin BAC = 10 / sin 45 -> CD = (10 / sin 45) * cos (a +45) ; CD =6 / cos a 6/ cos a = 10 * sqrt 2 * ( cos a * sqrt 2 /2 - sin a * sqrt 2 /2) -> 3 / cos a = 5 (cos a - sin a) -> 3/5 = cos^2 a - sin a * cos a -> 3/5 (cos^2 a + sin^2 a) = cos^2 a - sin a *cos a ; divide by cos ^2 a -> 3/5 (1 + tan^2 a) = 1 - tan a ; multiply by 5 -> 3 tan^2 a + 5 tan a - 2 = 0 ; solve the equation, tan a = -2 or tan a = 1/3 a is betwen 0 and 90 degrees so tan a is positive, so tan a = 1/3 DB/6 = tan a -> DB = 6* 1/3 -> DB = 2 ...

[@PreMath]

Thanks ❤️

[@giuseppemalaguti435]

Posto a=DB...arctg6/(10+a)+45+arctg a/6=90...applico tg..a=2..P=6+12+√180=18+√180

[@PreMath]

Excellent! Thanks ❤️

[@davoodzamani4254]

Thanks for video

[@PreMath]

You are very welcome! Thanks Davood❤️

[@waheisel]

I got x^2+10x-24=0 by placing a point F on AB 6 units from B. Then triangle ACD is similar to CFD. From that you get CD^2=60-10x. And from applying Pythagoras to BCD you get CD^2=x^2+36. Combine those to get x^2+10x-24=0. Thanks again PreMath for the fun puzzle.

[@holyshit922]

Playing with tangents we can calculate length of DB From Pythagorean theorem we can get length of AC Finally we add them

[@PreMath]

Thanks ❤️

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال a قياس الزاوية CAD وBD=y tana=6/10+y tan(a+45)=6/y y=2 AC=6(جذر5) محيط ABC هو (جذر5) + 18

[@prossvay8744]

Perimeter of the green triangle=18+6√5=31.42 units. ❤❤❤ Thanks sir.

[@PreMath]

Very good You are very welcome! Thanks ❤️

[@vaggelissmyrniotis2194]

I had to solve the equation x^4+20*x^3+172*x^2+720*x-2304=0 where x=BD in my own way but the channel's solution is way more clever and faster.

[@SanjaySinghaniaIN]

I'd some quick calculations with letting angle DCB be x => angle CDB = 90 - x => angle ADC = 90 + x => angle CAD = 45 - x. Now if DB be 'd', then tan x = d/6. And by tan(angle CAD) = 6/(10+d) => tan(45 - x) = 6/(10+d); this gives a quadratic in d: d^2 + 10d - 24 = 0 => d is either 2 or -12. Obviously d = 2, which makes sides as 6, 6sqrt(5) and 12. Thus ans = 6(sqrt(5) + 3).

[@PreMath]

Thanks ❤️

[@marcgriselhubert3915]

Let's name U = angle DCB. In right triangle DCB we have DB = BC.tan(U) = 6.tan(U), and in right triangle ABC we have AB = BC.tan(U + 45°) = 6.tan(U + 45°) So, AB = 6.((1 + tan(U)/(1 -tan(U)) using the formula giving tan(a+b). Let's name x = tan(U), using AB = AD + DB = 10 + DB we then have: 6.((1+x)/(1-x)) = 10 + 6.x or (10 + 6.x).(1-x) = 6.(1+x). We develop and obtain 6.(x^2) +10.x -4 = 0 or 3.(x^2) +5.x -2 = 0 Delta = 25 + 24 = 49, so x = (-5+7)/6 = 1/3 or x = (-5-7)/2 = -2 which is rejected as beeing negative. So x = tan(U) = 1/3. Now we have DB = 6.tan(U) = 6.(1/3) = 2 and AB = 10 + 2 = 12 The Pytagorean theorem gives then in triangle ABC that AC^2 = AB^2 + AC^2 = 6^2 + 12^2 = 36 + 144 = 180, and AC = sqrt(180) = 6.sqrt(5) Finally the perimeter of the green triangle ABC is AB + AC +BC = 12 + 6 + 6.sqrt(5) = 18 +6.sqrt(5)

[@PreMath]

Excellent! Thanks ❤️

[@ybodoN]

In summary: π − (tan⁻¹(x/6) + π/4 + π/2) = tan⁻¹(6/(10 + x)) ⇒ x = 2 Then we have AB = 12 and BC = 6 ⇒ AC = √(12² + 6²) = √180 = 6√5 The perimeter of the green triangle ABC is 12 + 6 + 6√5 = 6 (3 + √5)

[@PreMath]

Thanks ❤️

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال CAB=a وBD=x CDB=a+45 tan a=6/10+x tan (a+45) =6/x x^2+10x-24=0,x>0 x=2

[@josephsalinas6725]

Fiz utilizando a tangente: tg α = x/6 e tg(45 + α) = (10 + x) /6

[@keithwood6459]

If a PreMath video is less than 5 minutes, it's 50:50 I can solve it. It drops off fast from there.

[@PreMath]

Thanks ❤️

[@AdemolaAderibigbe-j8s]

Let < CAB = A then < CDB=A+45 (Sum of interior angles of triangle ACD) and < DCB = 45 - A (sum of angles of triangle CDB = 180 degrees), also < ACB = 45 +(45-A) = 90-A Let side DB =X 6/sin(A+45) = X/sin (45-A) (sin rule for triangle CDB) and 6/sin(A) = (10+X)/ sin(90-A) (for triangle ABC) (*) Substitute sin(A+45) = sin(A)cos(45) + cos(A)sin(45) and sin(45-A) = sin(45)cos(A) - cos(45)sin(A) and sin(45) = cos(45) = 1/(sqrt(2)) in the equations above to get 6(cos(A) - sin(A)) = X*(sin(A)+cos(A)) which is the same as cos(A)*(6 -X) = sin(A)*(X+6) and the second equation in line (*) is the same as cos(A)*6 = sin(A)*(10+X) Divide the two equations to get (6 - X)/6 = (X+6)/(10+X) Simplify the resulting equation to get X*X +10X -24 =0 or (X+12)(X-2) =0 and thus X = 2 (since X must be positive) Next use Pythagoras' theorem to find the hypotenuse of triangle ABC as square root of(12*12+6*6) or 6*sqrt(5) Thus perimeter of triangle ABC = 12+6 +sqrt(5)

[@PreMath]

Thanks ❤️

[@ultimategamer5843]

Sir how to solve this question: Show that 3x^10-y^10=1991 has no integral solution

[@ยี่สิบเก้าพฤศจิกา]

If X=2 : Area ABC = =(1/2)×12×6=36 และ BCD=(1/2)×2×6=6 แสดงว่า ACD=ABC-BCD=36-6=30

[@wackojacko3962]

The Solutions of a Quadratic Equation are all numbers (roots) which make the equation true. And all Quadratic Equations have 2 solutions (roots): Real and Distinct Real and Equal or Imaginary (Complex). So @ 2:59 we have a Dilemma and choose x=2 as more accurate than x=-12. I'm not gonna remain ignorant to some fundamental property of reality forever. I believe it exists. How much fun can one person have to start out the day without PreMath? 🙂

[@PreMath]

Thanks for the nice feedback❤️😀

[@murdock5537]

φ = 30°; ∆ ABC → AB = AD + BD = 10 + x; BC = 6; sin⁡(ABC) = 1; CAB = δ; BCD = α; DCA = 3φ/2; perimeter ∆ ABC = ? tan⁡(δ) = 6/(10 + x) → cot⁡(δ) = (10 + x)/6 = tan⁡(3φ/2 + α) = (tan⁡(3φ/2) + tan⁡(α))/(1 - tan⁡(3φ/2)tan⁡(α)) ↔ tan⁡(3φ/2) = 1 → tan⁡(3φ/2 + α) = (1 + tan⁡(α))/(1 - tan⁡(α)) → tan⁡(α) = x/6 → tan⁡(3φ/2 + α) = (6 + x)/(6 - x) = (10 + x)/6 → x^2 + 10x = 24 → x1 = 2; x2 = -12 < 0 ≠ solution → perimeter ∆ ABC = 12 + 6 + √(144 + 36) = 6(3 + √5)

[@PreMath]

Thanks ❤️

[@StephenRayWesley]

(10)^2=100 (6)^2=36 3x(15°)=45°x, 3x(15°)=45°x (45°x+45°x+90°)=180°x^2 (100+36)=136 (180°x^2-130)=√50°x^√2 2^√25 x^√2^1 2^5x^√1^√1 (x+2x-5)

[@PreMath]

Thanks ❤️