Solution using trigonometry: Applying the Pythagorean theorem, we find length BC = 40. Drop a perpendicular from D to AB and label the intersection E. ΔBDE is similar to ΔABC. DE = (24/40)x = 3x/5 and BE = (32/40)x = 4x/5. tan(2α) = 24/32 = 3/4. Applying the tangent double angle formula, tan(2α) = (2tan(α))/(1 - tan²(α)), we find tan(α) = 1/3, so DE/AE = 1/3, (3x/5)/AE = 1/3 and AE = 9x/5. However, BE + AE = AB = 32, so 4x/5 + 9x/5 = 13x/5 = 32 and x = 160/13 units, as PreMath also found.
In ∆ ABC BC^2=AB^2+AC^2 BC^2=32^2+24^2 BC=40 units Let E middle BC AE=BE=CE=20 LBAD=LEAD=a Let alpha=a) 32/20=x/20-x So x=160/13=12.31 units.❤❤❤ Thanks sir.
In ABD sin 180-3a/32=sin a/x sin 3a/32=sin a/x (3sin a-4sin^3 a)/32=sin a/x 3-4sin^2 a=32/x x= 34 /3-4 sin^2 a -# In ABC cos 2a = 32/40 = 4/5 We know that (1-cos 2a)/2 = sin^2 a So (1-4/5)*1/2= sin^2 a 1/10= sin^2 a We have x= 34 /3-4 sin^2 a So x=34/(3-4/10) x=34/(26/10) x= 34*10/26 x=170/13
The whole hypotenuse is 40 due to the triangle being a multiple of 3,4,5. Angle B (2 alpha) is tan(-1)(3/4) because 3/4 is 24/32 reduced. 36.87 degrees. Therefore, alpha is 18.435 degrees. The third angle will be 124.865 degrees. Length 32 is given, so x can be calculated by employing the law of sines, but as it's 11 at night I won't bother :) .
@ 3:13 I think we should draw the median from the right angle first and then consider a triangle BPA and see that it is iscoceles because the median is the radius of a circle drawn arond the original triangle. Only after that we can see the equal angles of 2alfa and make a conclusion that we can use a bicector ratio.
By observation, as CA = 24 = 3(8) and AB = 32 = 4(8), ∆CAB is an 8:1 ratio 3-4-5 Pythagorean triple right triangle and BC = 5(8) = 40. Draw DE, where E is the point on AB where DE is perpendicular to AB. As ∠DEB ~ ∠CAB and ∠EBD = ∠ABC, ∆DEB and ∆CAB are similar. As we know the dimensions of triangle ∆CAB, we can determine the trig values for 2α directly, and from there use known double angle identities to determine α. sin(2α) = 24/50 = 3/5 cos(2α) = 32/40 = 4/5 tan(2α) = 24/32 = 3/4 Finding the tan of α seems like it would be the best option, as it would allow us to tie DE to AE, since similarity already gives us a ratio for DE to EB. tan(2α) = 2tan(α)/(1-tan²(α)) 3/4 = 2u/(1-u²)
Let's note t instead of alpha (easier to write). In triangle ABC we have tan(2.t) = 24/32 = 3/4, so if u = tan(t) then 3/4 = (2.u)/(1 -u^2) and 8.u = 3 - 3.u^2 or 3.u^2 +8.u -3 = 0. Deltaprime = 16 +9 = 25 = 5^2, so u = (-4-5)/3 = -3 which is rejected as negative, or u = ((-4+5)/3 = 1/3. So tan(t) = 1/3 Now let H be the orthogonal projection of D on (AB). We have AH = HD/tan(t) =3.HD and BH = HD/tan(2.t) = (4/3).HD As AH + BH = AB = 32, xe then have that 3.HD + (4/3).HD = 32, or (13/3).HD = 32 which gives that HD = 96/13 and that BH = (4/3).96/13) = 128/13 We finish with the Pythagorean theorem in triangle BHD: BD^2 = HD^2 + BH^2 = (96/13)^2 + (128/13)^2 = (16384/169) + (9216/169) = 25600/169 and BD = sqrt(25600/169) = 160/13.
12.31 another method using trigonometry We know this is a 3-4-5 scaled up by 8. Hence, the angles 90, 36.87, and 53.13 degrees Hence, 2a = 36.87 since it faces 24 (8 *3), and a= 18.435 since it is 1/2 (2a). Hence, the other angle in triangle ABC = 124.695 [ 180 - (36.87) + 18.435] To find x using the law of sine a/sine a = b/sine b 32/sine 124.695 = x/ sine 18.435 32(sine 18.435/sine 124.695) =x 32 * 0.3846=x 12.307 =x 12.31 =x Answer
cos(2*alpha)= 4/5 e sen(2*alpha)=3/5 cos(alpha)=3/sqr(10) e sen(alpha)=1/sqr(10) sen(3*alpha)= 13/(5*sqr(10)) ABD x/sen(alpha)=32/sen(3*alpha)] x/1/sqr(10)=32/13/(5*sqr(10) x=32*5/13=160/13
Nice solution Prof! In mine I found tan 3alpha (Angle ADC is 2 alpha because external angle) once found tan 2 alpha and tan alpha. Then found the height AH = AB*AC/BC BH = AB²/BC (Euclid th.) X = BH - DH = AB²/BC - AH/tan 3alpha
Your construction of isosceles triangle ABP is an excellent example of a good construction. I started with a less smart but more intuitive construction with the constructed line AP being perpendicular to BC forming a right-angled triangle ABP and I got the answer quickly (important in exam) though in a less elegant way. 1. Get BC = 40 by Pythagoras theorem. 2. Get sin2a = 24/40 = 3/5, cos2a = 32/40 = 4/5 from the right-angled triangle ABC. 3. AP = 32 sin2a = 128/5 4. BP = 32 cos2a = 96/5 5. Get 2a from arcsin2a or arccos2a, then 3a = 3/2 (2a) = 55.3, then tan3a = 13/9 6. x = BP - AP/tan3a = 128/5 - (96/5)/(13/9) = 160/13
Drawing DQ, Q at AB with |DQ|=x, ∡BQD=2α, then ∡QDE=α, |QA|=x, |BQ|=32-x, draw also RD height of BQD with R at BQ, then |BR|=(32-x)/2=16-x/2. ◺BRD is similar to ◺ABC( 2α common, 90º), then at ◺ABC: cos(2α)=32/40=4/5, at ◺BRD: cos(2α) (16-x/2)/x=16/x-1/2. Then 16/x-1/2=4/5 so 16/x=4/5+1/2=13/10, then x/16=10/13; finally x=160/13.
Well, I need to watch because I don't know what the angle bisector theory is - but I do know x is about 12.3. Here's how. The right triangle is a 3-4-5, scaled by 8, so BC is 40. 2𝜽 is arcsin(24/40), or about 36.9 degrees. 𝜽 is ~18.4 degrees. With some fog from rounding, the angle at D is 125. By the law of sines, 32/sin(125) equals x/sin(18.4), so 32sin(18.4)/sin(125) equals x.
Hi i'm an 15 years old boy from italy and i always watch your video; they ate fandastic and very well made. I and one of my friend are going to attempt IMO at the italian final stage and we'd like to write a book with lots of math exercises in preparaction yo Cesenatico (italian stage).We think that the ezercise you post on RUclips fit what are we going to do. So,we are asking if for you is okey id we use some of your exercise and if you'd like to revise the argomentsof the soluction we are going to write based on the video.and we'd also like yo know if the exercise are taken form some olimpiad book and if there is copyright on them.
BC=40, we can calculate tan 2a=3/4, so, find out a, and then , using sine rule, we can directly compute x😅.It is a clumsy way, I do not do exact computation. 😂
Sorry to say, your method is rather clumsy, I can't understand why abandon the old trick double-angle- trick, angle-bisector trick make you run a longer route😢😢😢
Thank you! Enjoyed the creation of Isosceles Triangles solving the problem.
Wonderful!
You are very welcome!
Thanks ❤️
Solution using trigonometry: Applying the Pythagorean theorem, we find length BC = 40. Drop a perpendicular from D to AB and label the intersection E. ΔBDE is similar to ΔABC. DE = (24/40)x = 3x/5 and BE = (32/40)x = 4x/5. tan(2α) = 24/32 = 3/4. Applying the tangent double angle formula, tan(2α) = (2tan(α))/(1 - tan²(α)), we find tan(α) = 1/3, so DE/AE = 1/3, (3x/5)/AE = 1/3 and AE = 9x/5. However, BE + AE = AB = 32, so 4x/5 + 9x/5 = 13x/5 = 32 and x = 160/13 units, as PreMath also found.
My solution was nearly identical, I used the tangent half angle theorem. That gives tangent theta equals 1/3. Then EA equals 9x/5
In ∆ ABC
BC^2=AB^2+AC^2
BC^2=32^2+24^2
BC=40 units
Let E middle BC
AE=BE=CE=20
LBAD=LEAD=a Let alpha=a)
32/20=x/20-x
So x=160/13=12.31 units.❤❤❤ Thanks sir.
You are very welcome!
Thanks for sharing ❤️
..tg2α=3/4..sec2α=5/4..32/sin3α=x/sinα..x=32/(3-4(sinα)^2)=32/(4(cosα)^2-1)=32/(2cos2α+1)=32/(2/5/4+1)=32/(13/5)=160/13
Excellent!
Thanks for sharing ❤️
Let's find x:
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Since ABC is a right triangle, we can apply the Pythagorean theorem to calculate BC. Afterwards we use this result to calculate sin(2α) and cos(2α):
BC² = AB² + AC² = 32² + 24² = 8²*4² + 8²*3² = 8²*5² = 40² ⇒ BC = 40
sin(2α) = AC/BC = 24/40 = 3/5
cos(2α) = √[1 − sin²(2α)] = √[1 − (3/5)²] = √(1 − 9/25) = √(25/25 − 9/25) = √(16/25) = 4/5
Now we can apply the law of sines to the triangle ABD:
sin(∠ADB)/AB = sin(∠BAD)/BD
sin(180° − ∠ABD − ∠BAD)/AB = sin(∠BAD)/BD
sin(180° − 2α − α)/32 = sin(α)/x
sin(180° − 3α)/32 = sin(α)/x
sin(3α)/32 = sin(α)/x
sin(3α)/sin(α) = 32/x
[sin(2α)cos(α) + cos(2α)sin(α)]/sin(α) = 32/x
[(3/5)*cos(α) + (4/5)*sin(α)]/sin(α) = 32/x
(3/5)*[1/tan(α)] + 4/5 = 32/x
From the right triangle ABC we can directly conclude that tan(2α)=AC/AB=24/32=3/4. Now we can proceed:
tan(2α) = 2*tan(α)/[1 − tan²(α)]
3/4 = 2*tan(α)/[1 − tan²(α)]
1 − tan²(α) = (8/3)*tan(α)
0 = tan²(α) + (8/3)*tan(α) − 1
tan(α) = −4/3 ± √[(4/3)² + 1] = −4/3 ± √(16/9 + 1) = −4/3 ± √(16/9 + 9/9) = −4/3 ± √(25/9) = −4/3 ± 5/3
Since α
Excellent!
Thanks for sharing ❤️
In ABD
sin 180-3a/32=sin a/x
sin 3a/32=sin a/x
(3sin a-4sin^3 a)/32=sin a/x
3-4sin^2 a=32/x
x= 34 /3-4 sin^2 a -#
In ABC
cos 2a = 32/40 = 4/5
We know that (1-cos 2a)/2 = sin^2 a
So (1-4/5)*1/2= sin^2 a
1/10= sin^2 a
We have
x= 34 /3-4 sin^2 a
So x=34/(3-4/10)
x=34/(26/10)
x= 34*10/26
x=170/13
The whole hypotenuse is 40 due to the triangle being a multiple of 3,4,5.
Angle B (2 alpha) is tan(-1)(3/4) because 3/4 is 24/32 reduced.
36.87 degrees. Therefore, alpha is 18.435 degrees. The third angle will be 124.865 degrees.
Length 32 is given, so x can be calculated by employing the law of sines, but as it's 11 at night I won't bother :) .
@ 3:13 I think we should draw the median from the right angle first and then consider a triangle BPA and see that it is iscoceles because the median is the radius of a circle drawn arond the original triangle. Only after that we can see the equal angles of 2alfa and make a conclusion that we can use a bicector ratio.
By observation, as CA = 24 = 3(8) and AB = 32 = 4(8), ∆CAB is an 8:1 ratio 3-4-5 Pythagorean triple right triangle and BC = 5(8) = 40.
Draw DE, where E is the point on AB where DE is perpendicular to AB. As ∠DEB ~ ∠CAB and ∠EBD = ∠ABC, ∆DEB and ∆CAB are similar. As we know the dimensions of triangle ∆CAB, we can determine the trig values for 2α directly, and from there use known double angle identities to determine α.
sin(2α) = 24/50 = 3/5
cos(2α) = 32/40 = 4/5
tan(2α) = 24/32 = 3/4
Finding the tan of α seems like it would be the best option, as it would allow us to tie DE to AE, since similarity already gives us a ratio for DE to EB.
tan(2α) = 2tan(α)/(1-tan²(α))
3/4 = 2u/(1-u²)
Let's note t instead of alpha (easier to write). In triangle ABC we have tan(2.t) = 24/32 = 3/4, so if u = tan(t) then 3/4 = (2.u)/(1 -u^2) and 8.u = 3 - 3.u^2
or 3.u^2 +8.u -3 = 0. Deltaprime = 16 +9 = 25 = 5^2, so u = (-4-5)/3 = -3 which is rejected as negative, or u = ((-4+5)/3 = 1/3. So tan(t) = 1/3
Now let H be the orthogonal projection of D on (AB). We have AH = HD/tan(t) =3.HD and BH = HD/tan(2.t) = (4/3).HD
As AH + BH = AB = 32, xe then have that 3.HD + (4/3).HD = 32, or (13/3).HD = 32 which gives that HD = 96/13 and that BH = (4/3).96/13) = 128/13
We finish with the Pythagorean theorem in triangle BHD: BD^2 = HD^2 + BH^2 = (96/13)^2 + (128/13)^2 = (16384/169) + (9216/169) = 25600/169
and BD = sqrt(25600/169) = 160/13.
Veri nice sharing❤❤❤❤❤
I solved it using the theorem of sines.
a= arctag(24/32)/2= arctan(3/4)/2;
x/sin(a)=32/sin(3a);
x=32sin(a)/sin(3a)= 12.30769
12.31
another method using trigonometry
We know this is a 3-4-5 scaled up by 8. Hence, the angles 90, 36.87, and 53.13 degrees
Hence, 2a = 36.87 since it faces 24 (8 *3), and a= 18.435 since it is 1/2 (2a).
Hence, the other angle in triangle ABC = 124.695 [ 180 - (36.87) + 18.435]
To find x using the law of sine
a/sine a = b/sine b
32/sine 124.695 = x/ sine 18.435
32(sine 18.435/sine 124.695) =x
32 * 0.3846=x
12.307 =x
12.31 =x Answer
cos(2*alpha)= 4/5 e sen(2*alpha)=3/5
cos(alpha)=3/sqr(10) e sen(alpha)=1/sqr(10)
sen(3*alpha)= 13/(5*sqr(10))
ABD x/sen(alpha)=32/sen(3*alpha)]
x/1/sqr(10)=32/13/(5*sqr(10)
x=32*5/13=160/13
Nice solution Prof!
In mine I found tan 3alpha (Angle ADC is 2 alpha because external angle) once found tan 2 alpha and tan alpha. Then found the height AH = AB*AC/BC
BH = AB²/BC (Euclid th.)
X = BH - DH = AB²/BC - AH/tan 3alpha
Your construction of isosceles triangle ABP is an excellent example of a good construction.
I started with a less smart but more intuitive construction with the constructed line AP being perpendicular to BC forming a right-angled triangle ABP and I got the answer quickly (important in exam) though in a less elegant way.
1. Get BC = 40 by Pythagoras theorem.
2. Get sin2a = 24/40 = 3/5, cos2a = 32/40 = 4/5 from the right-angled triangle ABC.
3. AP = 32 sin2a = 128/5
4. BP = 32 cos2a = 96/5
5. Get 2a from arcsin2a or arccos2a, then 3a = 3/2 (2a) = 55.3, then tan3a = 13/9
6. x = BP - AP/tan3a = 128/5 - (96/5)/(13/9) = 160/13
AP=96/5.
Nice method.
Many many thanks❤️
Drawing DQ, Q at AB with |DQ|=x, ∡BQD=2α, then ∡QDE=α, |QA|=x, |BQ|=32-x, draw also RD height of BQD with R at BQ, then |BR|=(32-x)/2=16-x/2. ◺BRD is similar to ◺ABC( 2α common, 90º), then at ◺ABC: cos(2α)=32/40=4/5, at ◺BRD: cos(2α) (16-x/2)/x=16/x-1/2. Then 16/x-1/2=4/5 so 16/x=4/5+1/2=13/10, then x/16=10/13; finally x=160/13.
Well, I need to watch because I don't know what the angle bisector theory is - but I do know x is about 12.3. Here's how.
The right triangle is a 3-4-5, scaled by 8, so BC is 40.
2𝜽 is arcsin(24/40), or about 36.9 degrees. 𝜽 is ~18.4 degrees. With some fog from rounding, the angle at D is 125.
By the law of sines, 32/sin(125) equals x/sin(18.4), so 32sin(18.4)/sin(125) equals x.
Now, I attempt again 😅, using double angle trick, we have ((32-x)/2)/x=4/5, 8x=5(32-x), 13x=160, x=160/13.🤗
,شكرا لكم على المجهودات
يمكن استعمال
cosABD=3/5
sinABD=4/5
cosBAD=3/(racine10)
SinBAD=1/(racine10)
sinBAD /x = sinADB/32
........
x=160/13
Hi i'm an 15 years old boy from italy and i always watch your video; they ate fandastic and very well made.
I and one of my friend are going to attempt IMO at the italian final stage and we'd like to write a book with lots of math exercises in preparaction yo Cesenatico (italian stage).We think that the ezercise you post on RUclips fit what are we going to do.
So,we are asking if for you is okey id we use some of your exercise and if you'd like to revise the argomentsof the soluction we are going to write based on the video.and we'd also like yo know if the exercise are taken form some olimpiad book and if there is copyright on them.
Fiz utilizando o triângulo ABD.
asnwer=5/3
Is it possible to solve with a perpendicular from A to BC?
Fun problem, but can you please avoid leaving the law in the title? It spoils the solution and makes the question less interesting.
I hadn't noticed it in the title until I read your comment. But I agree with you.
12.307 or 12.31
BC=40, we can calculate tan 2a=3/4, so, find out a, and then , using sine rule, we can directly compute x😅.It is a clumsy way, I do not do exact computation. 😂
Thanks for sharing ❤️
Sorry to say, your method is rather clumsy, I can't understand why abandon the old trick double-angle- trick, angle-bisector trick make you run a longer route😢😢😢