Extend DO 3 units to E. Extend the arc of the quarter circle around to E to create a semicircle. Extend CB to E. as D and E are ends of a diameter and ∠C = 90, CB is colinear with CE. Triangle ∆DCE: DC² + CE² = ED² 2² + CE² = 6² CE² = 36 - 4 = 32 CE = √32 = 4√2 As ∠EOB = ∠DCE = 90° and ∠BEO = ∠CED, ∆EOB and ∆DCE are similar. OB/OE = DC/CE OB/3 = 2/4√2 = 1/2√2 OB = 3/2√2 Let P be the center of the circle, T be the point of tangency between circle P and the circumference of quarter circle O, and V be the point of tangency between circle P and CB. Draw PA, PV, and PB. AB and VB are tangents from circle P intersecting at B, so AB = VB. As PA = PV = r, and PB is shared, ∆PAB and ∆BVP are congruent. Let ∠ABP = ∠PBV = α. As ∠ABV thus equals 2α, and OA and VE are straight lines, ∠OBE = 2α as well, being a vertical angle of ∠ABV. As ∆EOB and ∆DCE are similar, ∠EDC = ∠OBE = 2α. tan(2α) = CE/DC = 4√2/2 = 2√2 sin(2α) = CE/ED = 4√2/6 = 2√2/3 cos(2α) = DC/ED = 2/6 = 1/3 tan(2α/2) = sin(2α)/(1+cos(2α)) tan(α) = (2√2/3)/(1+1/3) tan(α) = (2√2/3)/(4/3) tan(α) = 2√2/4 = 1/√2 Triangle ∆PAB: tan(α) = PA/AB 1/√2 = r/AB AB = √2r OA = AB + OB OA = √2r + 3/2√2 OA = (4r+3)/2√2 Draw radius OT. As the centers of two circles are colinear with their point of tangency, OT passes through P. As PT = r and OT = 3, OP = 3-r. Triangle ∆PAO: PA² + OA² = OP² r² + ((4r+3)/2√2)² = (3-r)² r² + (16r²+24r+9)/8 = 9 - 6r + r² 16r² + 24r + 9 = 72 - 48r 16r² + 72r - 63 = 0 16r² + 84r - 12r - 63 = 0 4r(4r+21) - 3(4r+21) = 0 (4r+21)(4r-3) = 0 r = -21/4 ❌ (r ≥ 0) | r = 3/4 ✓
Being 'α' the angle between OD and CB, and also between OD and BD , this means α = Angle BDO sin α = 2 / 6 = 1/3 Also α is angle POA, due to triangle BDO is similar to triangle POA r / (R-r) = 1/3 3r = R-r 4r = R r = ¼R = 3/4 = 0.75 cm ( Solved √ )
@@zsboya As you can see, the result is correct, r = 0.75 cm , same result as video and other comments. Those triangles are similar !! However, It seems that this exercise was created somewhat misleading. Those triangles are similar only with these dimensions, both angles have same value, but they DON'T HAVE TO BE. So, you are right !!! I was wrong !!! I fell into the trap !! See below other solution:
Labelling OB=a, BD=b sin α = 2/6 α = 19,4712° b = 3/ cos α b = 3,18198 cm Similarity of triangles: a/b = 2/6 a = 1,06066 cm α + 2θ = 90° θ = 35,2644° Tangent secant theorem: (a + r/tanθ)² = R. (R-2r) (a + r/tanθ)² = 3(3-2r) (a + r/tanθ)² = 9-6r Clearing 'r' : r = 0.75 cm ( Solved √ )
I think it is likely that this would be a simple proof of the task, only that it would be necessary to somehow prove that the triangles CDE and APO are similar.
Btw - your solution is excellent. You may be able to shorten it a bit by bisecting OC (C’ as others suggest), then finding the angle CDO (2 theta) by solving the triangle C’DO, as OC’D must be a right angle. Cos(2 theta) = 1/3 or OC’ = root(9-1) = 2*root(2), so tan(2 theta) = 2*root(2).
sin α = ½ 2 / 3 = 1/3 This right triangle will be named ODC' C' is midpoint of CD, and C'D=1 Rotate triangle ODC' about O, until align OD and OP: sin α = 1/3 = r / (R-r) 3.r = R-r 4r = R r = R/4 = 3/4 = 0.75 cm (.Solved √ )
@@jarikosonen4079 Very challenging this exercise of Math Booster. Nowadays, the prove is drawing to scale, in computer, in CAD. Enough provement !! I'm engineer, not mathematician
If you rotate triangle OPA about O so that OP aligns to OD, it's obvious that triangles OPA is similar to triangle FDC. So OP/PA = FD/CD --> (3-r)/r = 6/2 --> 3-r = 3r --> 3 = 4r --> r = 3/4.
Nice But you need to prove that the two triangles are similar. One angle is 90° that is equal in both triangles so you need to show that any second angle is also equal in those two triangles.
Not gonna lie, this one I had to solve in a calculator. And it took me about 2 hours (!!!) to properly set up the problem. As usual I only check the solution in the video after I found the answer myself, and it turns out the method by Math booster is similar to what I used, but her method is easier computation-wise.
As usual I want to try to avoid trigonometry. We assume and accept the fact that: BO=(3/4)√2, CE=4√2, CDE=2ϑ, ABP=ϑ. Now draw from O the perpendicular to the chord CE, which intersects CE itself at R and the (big) circumference at H. We observe that ORE is similar to DCE with a similarity ratio of 1:2, so that RO=(1/2)CD=1. Consequently HR=HO-RO = 3-1=2. Since the perpendicular to the chord through the center divides the chord itself into two equal parts, CR=RE and therefore RE=(1/2)CE)=2√2. Applying the Pythagorean theorem to the triangle HRE we have: HE=√(HR²+RE²)=√(4+8)=2√3.Applying the Pythagorean theorem again to the triangle DHE we have: HD=√(DE²-HE²)= √(36-12)=2√2√3. The ratio HE:HD is therefore 2√3:2√3√2= √2/2. We observe that the two arcs HE and HC are equal and therefore subtend equal angles to the circumference, so (in view of the fact that the total arc CE subtends CDE=2ϑ) HDE=HCD=ϑ. Consequently the right triangles DHE and PAB are similar and therefore HE:HD=PA:AB=r/AB=√2/2 for which AB=r√2. Tracing the ray from O passing through P until it meets the greater circumference in M and the smaller in G, we can write OA²=OM OG --> (AB+BO)²=OM OG --> (r√2+ (3/4)√2)² = 3(3-2r) --> ... 16r²+72r-63=0, which finally yields r=3/4.
Once we've constructed △CDE (as shown at 5:56) we can find CE using Pythagorean Theorem:
CE = √(DE² − CD²) = √(6² − 2²) = √32 = 4√2
sin 2θ = CE/DE = (4√2)/6 = 2√2/3
cos 2θ = CD/DE = 2/6 = 1/3
tan 2θ = CE/CD = (4√2)/2 = 2√2
We can get tan θ using half-angle identity:
tan θ = (1 − cos 2θ) / sin 2θ = (1 − 1/3) / (2√2/3) = (3−1)/(2√2) = 1/√2
In △OBE
tan2θ = OE/OB → OB = OE/tan2θ = 3/(2√2)
In △ABP
tan θ = AP/AB → AB = AP/tan θ = r/(1/√2) = r√2
In △OAP
AP = r
OA = OB + AB = 3/(2√2) + r√2 = (3+4r)/(2√2)
OP = 3−r
Using Pythagorean theorem, we get:
r² + ((3+4r)/(2√2))² = (3−r)²
r² + (16r²+24r+9)/8 = 9 − 6r+ r²
(16r²+24r+9)/8 = 9 − 6r
16r² + 24r + 9 = 72 − 48r
16r² + 72r − 63 = 0
16r² − 12r + 84r − 63 = 0
4r (4r − 3) + 21 (4r − 3) = 0
(4r + 21) (4r − 3) = 0
Since r > 0, then
*r = 3/4*
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
We need to prove that the right triangles CDE and APO are similar. In this case, we can calculate "r" very easily with ratio pairs. But how?
\ a =
After a little head scratching …. Elegantly done.
Extend DO 3 units to E. Extend the arc of the quarter circle around to E to create a semicircle. Extend CB to E. as D and E are ends of a diameter and ∠C = 90, CB is colinear with CE.
Triangle ∆DCE:
DC² + CE² = ED²
2² + CE² = 6²
CE² = 36 - 4 = 32
CE = √32 = 4√2
As ∠EOB = ∠DCE = 90° and ∠BEO = ∠CED, ∆EOB and ∆DCE are similar.
OB/OE = DC/CE
OB/3 = 2/4√2 = 1/2√2
OB = 3/2√2
Let P be the center of the circle, T be the point of tangency between circle P and the circumference of quarter circle O, and V be the point of tangency between circle P and CB. Draw PA, PV, and PB. AB and VB are tangents from circle P intersecting at B, so AB = VB. As PA = PV = r, and PB is shared, ∆PAB and ∆BVP are congruent.
Let ∠ABP = ∠PBV = α. As ∠ABV thus equals 2α, and OA and VE are straight lines, ∠OBE = 2α as well, being a vertical angle of ∠ABV. As ∆EOB and ∆DCE are similar, ∠EDC = ∠OBE = 2α.
tan(2α) = CE/DC = 4√2/2 = 2√2
sin(2α) = CE/ED = 4√2/6 = 2√2/3
cos(2α) = DC/ED = 2/6 = 1/3
tan(2α/2) = sin(2α)/(1+cos(2α))
tan(α) = (2√2/3)/(1+1/3)
tan(α) = (2√2/3)/(4/3)
tan(α) = 2√2/4 = 1/√2
Triangle ∆PAB:
tan(α) = PA/AB
1/√2 = r/AB
AB = √2r
OA = AB + OB
OA = √2r + 3/2√2
OA = (4r+3)/2√2
Draw radius OT. As the centers of two circles are colinear with their point of tangency, OT passes through P. As PT = r and OT = 3, OP = 3-r.
Triangle ∆PAO:
PA² + OA² = OP²
r² + ((4r+3)/2√2)² = (3-r)²
r² + (16r²+24r+9)/8 = 9 - 6r + r²
16r² + 24r + 9 = 72 - 48r
16r² + 72r - 63 = 0
16r² + 84r - 12r - 63 = 0
4r(4r+21) - 3(4r+21) = 0
(4r+21)(4r-3) = 0
r = -21/4 ❌ (r ≥ 0) | r = 3/4 ✓
😅😅😮
Being 'α' the angle between OD and CB, and also between OD and BD , this means α = Angle BDO
sin α = 2 / 6 = 1/3
Also α is angle POA, due to
triangle BDO is similar to triangle POA
r / (R-r) = 1/3
3r = R-r
4r = R
r = ¼R = 3/4 = 0.75 cm ( Solved √ )
where is the proof ta triangle BDO IS SIMILAR TO TRIANGLE POA?
Hi!
BDO = BEO = α is understandable since triangle BDE is isosceles. But how does it follow that the triangles BDO and POA are similar?
@@zsboya
As you can see, the result is correct,
r = 0.75 cm , same result as video and other comments.
Those triangles are similar !!
However, It seems that this exercise was created somewhat misleading.
Those triangles are similar only with these dimensions, both angles have same value, but they DON'T HAVE TO BE.
So, you are right !!! I was wrong !!!
I fell into the trap !!
See below other solution:
Labelling OB=a, BD=b
sin α = 2/6
α = 19,4712°
b = 3/ cos α
b = 3,18198 cm
Similarity of triangles:
a/b = 2/6
a = 1,06066 cm
α + 2θ = 90°
θ = 35,2644°
Tangent secant theorem:
(a + r/tanθ)² = R. (R-2r)
(a + r/tanθ)² = 3(3-2r)
(a + r/tanθ)² = 9-6r
Clearing 'r' :
r = 0.75 cm ( Solved √ )
I think it is likely that this would be a simple proof of the task, only that it would be necessary to somehow prove that the triangles CDE and APO are similar.
Labelling OB=a, BD=b
sin α = 2/6
α = 19,4712°
b = 3/ cosα =
b = 3,18198 cm
Similarity of triangles:
a/b = 2/6
a = 1,06066 cm
α + 2θ = 90°
θ = 35,2644°
Tangent secant theorem:
(a + r/tanθ)² = R. (R-2r)
(a + r/tanθ)² = 3(3-2r)
(a + r/tanθ)² = 9-6r
Clearing 'r' :
r = 0.75 cm ( Solved √ )
Btw - your solution is excellent. You may be able to shorten it a bit by bisecting OC (C’ as others suggest), then finding the angle CDO (2 theta) by solving the triangle C’DO, as OC’D must be a right angle. Cos(2 theta) = 1/3 or OC’ = root(9-1) = 2*root(2), so tan(2 theta) = 2*root(2).
sin α = ½ 2 / 3 = 1/3
This right triangle will be named ODC'
C' is midpoint of CD, and C'D=1
Rotate triangle ODC' about O, until align OD and OP:
sin α = 1/3 = r / (R-r)
3.r = R-r
4r = R
r = R/4 = 3/4 = 0.75 cm (.Solved √ )
ODC' ~ EDC => DC'=DC/(6/3)=1... Then how the rest?
Not sure if this is proven.
@@jarikosonen4079
Very challenging this exercise of Math Booster.
Nowadays, the prove is drawing to scale, in computer, in CAD.
Enough provement !!
I'm engineer, not mathematician
@marioalb9726 It looks good to check it using actual drawing.
In the solution, you assumed that O,P,M reside on the SAME line OM. That enables you to conclude that OP=3-r
How can you prove that?
If you rotate triangle OPA about O so that OP aligns to OD, it's obvious that triangles OPA is similar to triangle FDC. So OP/PA = FD/CD --> (3-r)/r = 6/2 --> 3-r = 3r --> 3 = 4r --> r = 3/4.
Nice
But you need to prove that the two triangles are similar. One angle is 90° that is equal in both triangles so you need to show that any second angle is also equal in those two triangles.
@MathBooster you are right. I wonder if there is a simple way to show say angle AOP = angle DFC.
Not gonna lie, this one I had to solve in a calculator. And it took me about 2 hours (!!!) to properly set up the problem. As usual I only check the solution in the video after I found the answer myself, and it turns out the method by Math booster is similar to what I used, but her method is easier computation-wise.
As usual I want to try to avoid trigonometry. We assume and accept the fact that: BO=(3/4)√2, CE=4√2, CDE=2ϑ, ABP=ϑ. Now draw from O the perpendicular to the chord CE, which intersects CE itself at R and the (big) circumference at H. We observe that ORE is similar to DCE with a similarity ratio of 1:2, so that RO=(1/2)CD=1.
Consequently HR=HO-RO = 3-1=2. Since the perpendicular to the chord through the center divides the chord itself into two equal parts, CR=RE and therefore RE=(1/2)CE)=2√2. Applying the Pythagorean theorem to the triangle HRE we have: HE=√(HR²+RE²)=√(4+8)=2√3.Applying the Pythagorean theorem again to the triangle DHE we have: HD=√(DE²-HE²)= √(36-12)=2√2√3. The ratio HE:HD is therefore 2√3:2√3√2= √2/2. We observe that the two arcs HE and HC are equal and therefore subtend equal angles to the circumference, so (in view of the fact that the total arc CE subtends CDE=2ϑ) HDE=HCD=ϑ. Consequently the right triangles DHE and PAB are similar and therefore HE:HD=PA:AB=r/AB=√2/2 for which AB=r√2. Tracing the ray from O passing through P until it meets the greater circumference in M and the smaller in G, we can write OA²=OM OG --> (AB+BO)²=OM OG --> (r√2+ (3/4)√2)² = 3(3-2r) --> ... 16r²+72r-63=0, which finally yields r=3/4.
Hi! Nice derivation, but it's very long. By the way, you wrote it wrong in one place: HDE = HDC (Not HCD) = ϑ.
You are right. Thank you.@@zsboya
Very impressive solution
Στο σημείο που θέλετε να δείξετε ότι
Nice pro If any with easy soln please tell thank you sir❤
I think it is not clear semicircle DCE contains semiquarter circle
Nice problem sir
Thank you 🙂
Good
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .