Math Olympiad | Can you find Perimeter of the Green triangle ABC? |
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- Опубликовано: 18 сен 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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Solution by tangent double angle formula tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)). Let
Excellent!
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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
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Very good!
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Love the step by step detailed solution!! More of such problems please!!
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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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.
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Posto a=DB...arctg6/(10+a)+45+arctg a/6=90...applico tg..a=2..P=6+12+√180=18+√180
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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
...
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Thank you for the solution, adding the additional lines identifying the congruent triangles! Appreciated this experience!
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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.
Thanks for the videos. You are very helpful!
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Perimeter of the green triangle=18+6√5=31.42 units. ❤❤❤ Thanks sir.
Very good
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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)
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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)
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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)
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Great question professor!! 👍
Glad you liked it!
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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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(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)
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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
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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❤️
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
If a PreMath video is less than 5 minutes, it's 50:50 I can solve it. It drops off fast from there.
Thanks ❤️
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
Thanks for video
You are very welcome!
Thanks Davood❤️
Fiz utilizando a tangente: tg α = x/6 e tg(45 + α) = (10 + x) /6
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❤️😀
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.
A beautiful identity is hidden in this puzzle, 45°=arctan 1/2+arctan 1/3.🎉
True!
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Sir how to solve this question:
Show that 3x^10-y^10=1991 has no integral solution
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)).😅😅😅😅😅😅😅😅
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If X=2 : Area ABC = =(1/2)×12×6=36 และ BCD=(1/2)×2×6=6 แสดงว่า ACD=ABC-BCD=36-6=30