Another solution: draw the bisector of the external angle A and let E be its intersection with base BC. By external angle bisector theorem we get EB/EC=AB/AC or (8+EC)/EC=5/3, thus EC=12. Triangle DAE is right, its hypotenuse is DE=12+3=15 and its angle E is 90-60=30 degrees, therefore AD=15/2.
Interesting, thank you for the solutions. Another approach similar to your second one would be as follows. Choose point P on AB s.t. [angle BDP] = 60°, then [DP] = 3 as the triangles DCA and DPA are congruent. Then choose point Q on BD s.t. [angle DQP] = 60°. One than not only has that triangle PQD is equilateral but, in particular, that QP is parallel to DA. Thus the triangles BQP and BDA are similar, therefore [PQ]/[BQ] = [AD]/[BD] = 3/2 = x/5, which leads to x = 15/2 = 7.5.
I used the sine rule in triangle ABD to get cot theta in terms of AD. I used the sine rule in triangle ACD to get cot theta in terms of AD. Then equating the two expressions and cancelling out gave the length of AD quite nicely. Thank you for all your work.
Interesting solutions! Another solution is drawing a circumcircle of ABC. Let E be the intersection of the circumcircle of ABC and the extension of AD, and H be the perpendicular foot from E on BC. Then EBC = EAC = EAB = ECB, so EB = EC. It means HB = BC = 4, so HD = 1. From EDH = 60 and HD = 1, we get DE = 2. AD * DE = BD * CD, so AD * 2 = 3 * 5, which yields AD = 15/2
Le résultat permet de voir à quel point la représentation graphique est fausse. Avec AD=7,5 et l'angle de 60°, la projection du sommet A sur la base BC, est à 3,75 du point D. Difficile de voir la similitude des triangles.
Another approach would be to drop a perpendicular from A to BC and label the intersection as point E. We note that ΔADE is a special 30°-60-90° right triangle. Let DE = a, then AE = a√3 and AD = 2a. Then, CE = 3 - a, tan(Θ - 30°) = (3-a)/(a√3) and tan(Θ + 30°) = (a + 5)/(a√3). Using the tangent sum of angles formula, tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)), twice, noting that tan(30²) = 1/√3 and tan(-30²) = -1/√3, we get two expressions for tan(Θ) which can be set equal to each other and solved for a, which is doubled to find AD. This method will work if Θ < 30°, in which case a will be greater than 3, which is the case here (a = 3.75). Point C is to the left of point E. The diagram is misleading because point E appears that it would be between C and D.
There is a easy way and easy to use 1.from C to AB at E make AE=AC=3y then EB=2y 2.bottom area Tri ECB can be split into 3A and 5A 3.Tri ACE : Tri BCE = 2:3=8A : 12A 4.lenth AD =15/2
Very nice methods! But for me it's difficult to see if 2 triangles are similar. So in my solution I used the law of sines twice: 3/sinθ = 2a/sqrt(3) -> sinθ = 3*sqrt(3)/2a and 5/sinθ = 2b/sqrt(3) -> sinθ = 5*sqrt(3)/2b, where a and b are unknown sides of the triangle. So 3*sqrt(3)/2a = 5*sqrt(3)/2b -> b = 5a/3. Then I used the law of cosines for the bottom side of the triangle: 64 = 25a^2 / 9 + a^2 - 2*(5a^2 / 3)*cos2θ. cos2θ = 1 - 2*(sinθ)^2 = 1 - 27/2a^2, then I found a: a= 3*sqrt(19)/2 and used cosines law again for a: (3*sqrt(19)/2)^2 = x^2 + 9 - 2x*3*sin60, solved it and that gave me the same answer
Вообще-то эта задачка решается в лоб без применения интеллекта. В немного более общем виде, пусть отрезки стороны BC m и n (m > n), биссектриса L, тогда существует число x>1, что AB = mx BC = nx; длина биссектрисы L² = mnx² - mn; по теореме косинусов L² + n² - Ln = n²x²; => mnx² - mn + n² - n √(mn)√(x²-1) = n²x²; (m-n)(x²-1) = √(mn)√(x²-1); √(x²-1) = √(mn)/(m-n); L = mn/(m-n); при m = 5 n = 3 L = 15/2;
Thank you for all these challenges. A small problem to submit to you: 'rue du chat qui peche' is a narrow street in Paris. Two ladders are placed across the street, one 3 meters, the other 2 meters. They cross at a height of 1 meter. How wide is the street?
9 - y.y = BC.BC where B is at the base of 2metre ladder , A at the base of 3metre ladder , C at the top of the 3metre ladder, D at the top of the 2metre ladder, 4 - y.y = AD.AD where y = AB which is the width of naughty puss street. (Cat-fisher?) and E the point where the ladders cross if I have understood the question correctly. (BC-AD)(BC+AD) = 5 ( because ( 9 - y.y) - ( 4 - y.y) = BC.BC - AD.AD = 5 Another 2metre ladder is placed across the street from A to F where F is drectly below C. Now I see triangles CFG and ESA are congruent. G is where the long ladder passes between the tops of the two shorter ladders. and S is at street level immediately below E So CF is also 1 metre. BC-AD is is therefore 1 metre. However BC+ AD must be less than 5 metres because AC+BD = 5 metres. Can any one see where I went wrong?
CF is greater than ES , so those triangles are similar and CFG and ESA are not congruent. The width of the street is SQRT(AE.AE-1) +SQRT(BE.BE-1) which is not neat. Scale drawing with compasses and trial - and - improvement might get close. Or take a measuring tape to Paris!
Another solution: draw the bisector of the external angle A and let E be its intersection with base BC. By external angle bisector theorem we get EB/EC=AB/AC or (8+EC)/EC=5/3, thus EC=12. Triangle DAE is right, its hypotenuse is DE=12+3=15 and its angle E is 90-60=30 degrees, therefore AD=15/2.
Interesting, thank you for the solutions. Another approach similar to your second one would be as follows. Choose point P on AB s.t. [angle BDP] = 60°, then [DP] = 3 as the triangles DCA and DPA are congruent. Then choose point Q on BD s.t. [angle DQP] = 60°. One than not only has that triangle PQD is equilateral but, in particular, that QP is parallel to DA. Thus the triangles BQP and BDA are similar, therefore [PQ]/[BQ] = [AD]/[BD] = 3/2 = x/5, which leads to x = 15/2 = 7.5.
Brilliant. Well done.
Sehr schön .
I used the sine rule in triangle ABD to get cot theta in terms of AD.
I used the sine rule in triangle ACD to get cot theta in terms of AD. Then equating the two expressions and cancelling out gave the length of AD quite nicely. Thank you for all your work.
Interesting solutions!
Another solution is drawing a circumcircle of ABC. Let E be the intersection of the circumcircle of ABC and the extension of AD, and H be the perpendicular foot from E on BC.
Then EBC = EAC = EAB = ECB, so EB = EC. It means HB = BC = 4, so HD = 1. From EDH = 60 and HD = 1, we get DE = 2.
AD * DE = BD * CD, so AD * 2 = 3 * 5, which yields AD = 15/2
👍, but...
HB=HC=BC/2=4 !!!
@@rabotaakk-nw9nm Oh, thank you. Yes, it was a typo. 😅
Le résultat permet de voir à quel point la représentation graphique est fausse. Avec AD=7,5 et l'angle de 60°, la projection du sommet A sur la base BC, est à 3,75 du point D. Difficile de voir la similitude des triangles.
Another approach would be to drop a perpendicular from A to BC and label the intersection as point E. We note that ΔADE is a special 30°-60-90° right triangle. Let DE = a, then AE = a√3 and AD = 2a. Then, CE = 3 - a, tan(Θ - 30°) = (3-a)/(a√3) and tan(Θ + 30°) = (a + 5)/(a√3). Using the tangent sum of angles formula, tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)), twice, noting that tan(30²) = 1/√3 and tan(-30²) = -1/√3, we get two expressions for tan(Θ) which can be set equal to each other and solved for a, which is doubled to find AD. This method will work if Θ < 30°, in which case a will be greater than 3, which is the case here (a = 3.75). Point C is to the left of point E. The diagram is misleading because point E appears that it would be between C and D.
Die Größe BD = 5 spielt hier überhaupt keine Rolle und wird nicht verwendet ?
Thank you professor.2nd method is very cool!
There is a easy way and easy to use
1.from C to AB at E make AE=AC=3y then EB=2y
2.bottom area Tri ECB can be split into 3A and 5A
3.Tri ACE : Tri BCE = 2:3=8A : 12A
4.lenth AD =15/2
Very nice methods! But for me it's difficult to see if 2 triangles are similar. So in my solution I used the law of sines twice: 3/sinθ = 2a/sqrt(3) -> sinθ = 3*sqrt(3)/2a and 5/sinθ = 2b/sqrt(3) -> sinθ = 5*sqrt(3)/2b, where a and b are unknown sides of the triangle. So 3*sqrt(3)/2a = 5*sqrt(3)/2b -> b = 5a/3.
Then I used the law of cosines for the bottom side of the triangle: 64 = 25a^2 / 9 + a^2 - 2*(5a^2 / 3)*cos2θ.
cos2θ = 1 - 2*(sinθ)^2 = 1 - 27/2a^2, then I found a: a= 3*sqrt(19)/2 and used cosines law again for a: (3*sqrt(19)/2)^2 = x^2 + 9 - 2x*3*sin60, solved it and that gave me the same answer
Вообще-то эта задачка решается в лоб без применения интеллекта. В немного более общем виде, пусть отрезки стороны BC m и n (m > n), биссектриса L, тогда существует число x>1, что AB = mx BC = nx; длина биссектрисы L² = mnx² - mn; по теореме косинусов L² + n² - Ln = n²x²; => mnx² - mn + n² - n √(mn)√(x²-1) = n²x²; (m-n)(x²-1) = √(mn)√(x²-1); √(x²-1) = √(mn)/(m-n); L = mn/(m-n); при m = 5 n = 3 L = 15/2;
Nice! φ = 30°; ∆ ABC → AB = AD + BD = 5 + 3; BCD = DCA = θ; CDB = 2φ; DBC = γ; CD = k
AC = n; BC = m → n/5 = m/3 → 3n = 5m → n = 5m/3 → AC = 5x → BC = 3x →
15x^2 - 15 = k^2 → k = √(15(x^2 - 1)); sin(2φ) = √3/2 → cos(2φ) = sin(φ) = 1/2
sin(4φ) = sin(6φ - 4φ) = sin(2φ) = √3/2 → cos(4φ) = -cos(6φ - 4φ) = -cos(2φ) = -1/2
∆ ABC → 64 = 25x^2 + 9x^2 - 30x^2cos(2θ) → cos(2θ) = (17x^2 - 32)/15x^2
∆ BCD → 9 = k^2 - 9x^2 - 6kxcos(θ) → cos(θ) = 4(x^2 - 1)/x√15(x^2 - 1)) →
cos^2(θ) = 16(x^2 - 1)/(15x^2) → sin^2(θ) = 1 - cos^2(θ) = (16 - x^2 )/(15x^2 ) →
∆ BCD → sin(2φ)/3x = sin(θ)/3 → sin(θ) = √3/2x →
sin^2(θ) = 3/4x^2 → cos^2(θ) = 1 - sin^2(θ) = (4x^2 - 3)/4x^2 →
cos(2θ) = cos^2(θ) - sin^2(θ) = (2x - 3)/2x^2 = (17x^2 - 32)/15x^2 →
x = √19/2 → k = 15/2 → 3x = m = 3√19/2 → n = 5√19/2 →
cos(θ) = 4√19/19 → θ ≈ 23,41°→ cos(γ) = -√19/38 → γ ≈ 96,59°
Thank you for all these challenges. A small problem to submit to you: 'rue du chat qui peche' is a narrow street in Paris. Two ladders are placed across the street, one 3 meters, the other 2 meters. They cross at a height of 1 meter. How wide is the street?
9 - y.y = BC.BC where B is at the base of 2metre ladder , A at the base of 3metre ladder , C at the top of the 3metre ladder, D at the top of the 2metre ladder,
4 - y.y = AD.AD where y = AB which is the width of naughty puss street. (Cat-fisher?) and E the point where the ladders cross
if I have understood the question correctly.
(BC-AD)(BC+AD) = 5 ( because ( 9 - y.y) - ( 4 - y.y) = BC.BC - AD.AD = 5
Another 2metre ladder is placed across the street from A to F where F is drectly below C.
Now I see triangles CFG and ESA are congruent.
G is where the long ladder passes between the tops of the two shorter ladders. and S is at street level immediately below E
So CF is also 1 metre. BC-AD is is therefore 1 metre. However BC+ AD must be less than 5 metres because AC+BD = 5 metres. Can any one see where I went wrong?
CF is greater than ES , so those triangles are similar and CFG and ESA are not congruent.
The width of the street is SQRT(AE.AE-1) +SQRT(BE.BE-1) which is not neat.
Scale drawing with compasses and trial - and - improvement might get close. Or take a measuring tape to Paris!
AD=X
AB can't be X
AB=5k AC=3k
is better
Method 2 is lovely.
Muito bonita a questão!!!! ;)
Why do I Need AD^2?
Yeah,the 2nd method is excellent.
Man i find another away but i dont know its that right it
asnwer=5x
2nd method is good
21.66 min go for me .today 😮💨
性 😊