Math Olympiad | Can you find Perimeter of the Green triangle ABC? |
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- Опубліковано 2 жов 2024
- Learn how to find the Perimeter of the Green triangle ABC. Important Geometry and Algebra skills are also explained: Perimeter of a triangle; similar triangles; congruent triangles; Pythagorean Theorem; right triangles; isosceles triangles. Step-by-step tutorial by PreMath.com
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Math Olympiad | Can you find Perimeter of the Green triangle ABC? | #math #maths #geometry
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Like for preMath👍
Thanks dear ❤️
@@PreMath I appreciate it🥰
Solution by tangent double angle formula tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)). Let
Excellent!
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Excellent! Tangent summation formula. Best solution I have seen.
I did like this too. 👋
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!
Very good!
You are very welcome!
Thanks ❤️
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)
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If a PreMath video is less than 5 minutes, it's 50:50 I can solve it. It drops off fast from there.
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Perimeter of the green triangle=18+6√5=31.42 units. ❤❤❤ Thanks sir.
Very good
You are very welcome!
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Love the step by step detailed solution!! More of such problems please!!
A beautiful identity is hidden in this puzzle, 45°=arctan 1/2+arctan 1/3.🎉
True!
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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.
That is how I did it too.
Thanks ❤️
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
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
premath resistance to employ the means of trigonometry, not to mention, multiple angle formulas, but in this puzzle, the solution is simplified while applying such formula. 😮
I like trigonometry equally as well!
Thanks for the feedback❤️
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.
Thank you for the solution, adding the additional lines identifying the congruent triangles! Appreciated this experience!
Glad to hear that!
You are very welcome!
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Fiz utilizando a tangente: tg α = x/6 e tg(45 + α) = (10 + x) /6
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)
Thanks ❤️
Posto a=DB...arctg6/(10+a)+45+arctg a/6=90...applico tg..a=2..P=6+12+√180=18+√180
Excellent!
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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? 🙂
Thanks for the nice feedback❤️😀
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
...
Thanks ❤️
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)
Thanks ❤️
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).
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Thanks for the videos. You are very helpful!
Glad you like them!
You are very welcome!
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Playing with tangents we can calculate length of DB
From Pythagorean theorem we can get length of AC
Finally we add them
Thanks ❤️
(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)
Thanks ❤️
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.
Sir how to solve this question:
Show that 3x^10-y^10=1991 has no integral solution
Great question professor!! 👍
Glad you liked it!
Thanks ❤️
I repair my work, my way should work, 😅
DB=6tan a, AB=10+6tan a, tan (45+a)=(1+tan a)/(1-tan a)=(10+6tan a)/6=5/3+tan a, 3+3tan a=(1-tan a)(5+3tan a） 3tan^2 a+5tan a-2=0,
-2, not -5, it is my mistake in my first attempt, so (3tan a-1)(tan a+1)=0, reject negative root, tan a=1/3, DB=2, AB=12, thus AC is sqrt(6^2+12^2)=6sqrt(5), therefore the answer is 18+6sqrt(5)=6(3+sqrt(5)).😅😅😅😅😅😅😅😅
Thanks ❤️
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)
Thanks ❤️
Thanks for video
You are very welcome!
Thanks Davood❤️
If X=2 : Area ABC = =(1/2)×12×6=36 และ BCD=(1/2)×2×6=6 แสดงว่า ACD=ABC-BCD=36-6=30