Sin BAC = 3/5, cos BAC = 4/5 Draw lines from O to A and from Q to C. The lines bisects the angles A and C. tan BAC/2 = sin / 1+cos = 3/5 / (1+ 4/5) = 1/3 tan ACB/2 = 4/5 / (1+ 3/5) = 1/2 AE = r/ tan BAC/2 -> AE = 3r DC = r/ tan ACB/2 -> DC = 2r AC = 3r + 2r +2r = 35 -> r = 5
1/The triangle ABC is a 3x7-4x7-5x7 triple so, to make the calculation simpler: Assume that r= (radius of the yellow circles/7) in a 3--4-5 triangle. AO is the bisector of the angle CAB intersecting BC at point M. By using the bisector theorem: CM/AC=BM/AB---> CM/5=BM/4= (CM+BM)/(5+4)=3/9= 1/3 ----> BM=4/3 Consider the triangle AMB and AOF, they are similar so OF/BM= AF/AB= a/4----> r/(4/3)= a/4----> a= 3r 2/ Similarly, the bisector CQ of the angle ACB intersects AB at point N, we have b=2r 3/AC= a+2r+b = 3r+2r+ 2r=5----> 7r=5---> r=5/7 The radius of the yellow circle= 7r= 5 units
Area ABO triangle: 28r/2 Area BCQ triangle: 21r/2 Area AOQC trapeze: (2r+35)r/2 Height OBQ triangle: 84/5-r so the area OBQ triangle: (84/5-r)2r/2 The sum of these areas: 21x28/2 So r=5
I've found the solution summing the areas inside ABC Area(AOB)+area(CQB)+ Area trapezoid(ACQO) + Area(OQB) = Area(ABC) 28*r*1/2+21*r*1/2+(35+2r)*r*1/2+(84/5-r)*2r*1/2=21*28*1/2 14r+21/2r+35/2r+r^2+84/5r-r^2=294 R=5
I think you made this a bit more complicated than necessary. First, it helps to scale all the distances by 7 at the start, so the triangle sides are 3, 4 and 5. Then at your 8:27, we have: (4 - a - r)/2r = 4/5 (3 - b - r)/2r = 3/5 20 - 5a - 5r = 8r 15 - 5b - 5r = 6r Add these two equations, giving 35 - 5(a + b) - 10r = 14r Now from the length of the hypotenuse, 5 = a + b + 2r, i.e. a + b = 5 - 2r 35 - 5(5 - 2r) = 24r, 20 + 10r = 24r, 20 = 14r, r = 10/14 = 5/7. Scaling back, r = 5.
Another sol: Consider lime EO,OF,DQ,QG Since they are radius they are equal hence the line is angle bisector using half angle sine formula we can easily see that the hypotenuse of triangle is 7r and is 35 hence r=5
Really interetsing and I am now grasping the idea of using equations to solve geometrical unknowns. Nobody ever explained it to me this way before. I am good at maths, but a sucession of bad maths teachers is cumulative, I'm afraid.
Radius (R) equal to 5 Linear Units. First draw a Scalene Triangle [ABC] as here given, with sides 21 / 28 / 35, being the point A the Origin of the Coordinates (0 ; 0). AC = 35, because sqrt(21^2 + 28^2) = sqrt(441 + 784) = sqrt(1.225) = 35. Check!! Now, if I draw a a Line X = 7 I get a Square 21 by 21. The Center (interception of the 2 Diagonals) of this Square have coordinates (16 ; 12). Passing a Perpendicular Line to AC and in point (16 ; 12), the Center of the Square, we get a Triangle with Sides 15 / 20 / 25. The Area of this Triangle is A = (Base * height) / 2 = (25 * 12) / 2 = 300 / 2 = 150 sq un Now : (15R + 20R + 25R) / 2 = 150 60R / 2 = 150 30R = 150 R = 150 / 30 R = 5 Answer: The Radius is equal to 5 Linear Units.
As BC = 21 = 3(7), and AB = 28 = 4(7), ∆ABC is a 7:1 ratio Pythagorean triple triangle, and CA = 5(7) = 35. Draw radii OE, OF, QD, and QG. As radii, these segments are perpendicular to the tangents: AB to OF, BC to QG, and CA to QD and OE. Draw OQ. The quadrilateral OQDE is a rectangle, as OQ and DE are parallel and QD and OE are both perpendicular to DE. Thus OQ = DE = 2r. Let M be a point on BC where OM is parallel to AB amd perpendicular to BC. Let N be a point on AB where QN is parallel to BC and perpendicular to AB. Let the intersection point of OM and QN be P. Take note of trisngle ∆OPQ. All of its sides are parallel to the sides of ∆ABC, therefore all of its angles are the same, and the two triangles are similar. As OQ = 2r, OP = 4/5(2r) = 8r/5, and PQ = 3/5(2r) = 6r/5. By two tangent theorem, GC and CD are congruent, and EA and AF are congruent. Let CD = GC = x and AF = EA = y. By observation, AB = y + 8r/5 + r, BC = r + 6r/5 + x, and CA = x + 2r + y. As we have numerical values for all three sides, we should be able to solve for r. y + 8r/5 + r = 28 ---- (1) y = 28 - 13r/5 r + 6r/5 + x = 21 ---- (2) x = 21 - 11r/5 x + 2r + y = 35 ---- (3) 21 - 11r/5 + 2r + 28 - 13r/5 = 35 24r/5 - 2r = 49 - 35 = 14 14r/5 = 14 r = (5/14)14 = 5
AC = 35 (easy). Now, let's name t = angleBAC, we have tan(t) = 21/28 = 3/4, so tan(t/2) = 1/3 (easy to obtain with the formula giving tan(2.x)) So, a = AF = R/tan(t/2) = 3.R in triangle AFO. Now let's name t' = angleACB, t' = 90° -t, so t'/2 = 45° -t/2 and tan(t') = 1/2 (easy to obtain with the formula giving tan(a -b)) So b = CG = R/tan(t') = 2.R in triangle CQG. Now let'a use an orthonormam, center A, first axis (AB), second axis (AC). We have O(a;R) or O(3.R;R) and Q(28 -R; 21 -b) or Q(28 -R; 21 -2.R) Then we have VectorOQ(28 -4.R; 21 -3.R) and OQ^2 = (28 -4.R)^2 + (21 -3.R)^2 = 784 -224.R +16.R^2 +441 -126.R +9.R^2 = 25.R^2 -350.R +1225 Or OQ = 2.R and OQ^2 = 4.R^2, then we have the equation: 21.R^2 -350.R +1225 = 0. Deltaprime = (-175)^2 -21.1225 = 4900 = (70)^2 So: R = (175 + 70)/21 = 245/21 = 35/3 which is rejected as in this case AF = a= 35 > AB =28, or R = (175 - 70)/21 = 5 which is our solution.
AE=a,DC=b...(28-(a+r))^2+(21-(b+r))^2=(2r)^2....35=(a+r)+(b+r)...sostituisco risulta (28-(a+r))^2+((a+r)-14)^2=4r^2..(1eq)...2arctg(r/a)=arctg21/28...a=3r(2eq)...inserisco la 2)nella 1)...(28-4r)^2+(4r-14)^2=4r^2.svoogendo i calcoli rimane 32r^2-336r+980=4r^2....r^2-12r+35=0..r=6+1=7..r=6-1=5.…uhm..credo r=5, sia corretta
... Good day to you, Step(1): Triplet( 7*3 - 7*4 - 7*5 ) gives us I AC I = 35 u. ... I ED I = I OQ I = 2r ( r is radius circles ) ... I AE I = I AF I = X & I CD I = I CG I = Y ... so, I FB I = 28 - X & I GB I = 21 - Y ... Angle(BAC) = ARCTAN(3/4) ... Step(2): Creating inside right triangle(OQS) , where Angle(QSO) = 90 deg. ... Vertical distance V between Q and O ... I QS I = 21 - Y - r & Horizontal distance H between Q and O ... I OS I = 28 - X - r and Angle(SOQ) = Angle(BAC) = ARCTAN(3/4) ... last but not least I AC I = 35 = X + 2r + Y ... now TAN(SOQ) = TAN(BAC) = 3/4 = I QS I / I OS I ... 3/4 = ( 21 - Y - r ) / (28 - X - r ) ... after a few steps we obtain ... r = 3X - 4Y or 2r = 6X - 8Y & 2r = 35 - ( X + Y ) ... which gives us X - Y = 5 or Y = X - 5 .... so, r = 3X - 4(X - 5) = 20 - X .... finally ... TAN(OAF) = I OF I / I AF I = r / X = TAN(ARCTAN(3/4)/2) , so X = r / TAN(ARCTAN(3/4)/2) & recalling r = 20 - X ... after a few steps we obtain r = 20 / ( 1 + 1/TAN(ARCTAN(3/4)/2) ) = 20 / 4 = 5 u. ... r = 5 u. ... now it is time to watch your, I assume, much easier and clear explanation sir (lol), but reading other comments I guess I found the right answer for radius r in the end (lol) ... many thanks for your continuing interesting and fun geometry exercises ... best regards, Jan-W
Naturally … yet another way, but essentially the SAME way as PreMath (but EASY! if you're good at manipulating basic algebra) Draw line between centers of the circles. [1.1] center-to-center length = 2𝒓 Figure length of hypotenuse of big triangle [2.1] hypotenuse = √( 21² + 28² ) [2.2] hypotenuse = 35 This could also have been found by noting that 21 = 3×7 and 28 = 4×7, so the hypotenuse must be 5×7. Oh, well, I'm not intuitive enough to do it the 'easy way'. Anyhow, now draw 2 more lines from the centers of the circles. A vertical one on the right circle and a horizontal one from the left circle. They intersect, forming a similar triangle to the big triangle. Similar means 'proportionate to'. So… [3.1] horizontal = ⅘ center-to-center … from [1.1] [3.2] horizontal = ⅘ × 2𝒓 [3.3] horizontal = 8⁄5𝒓 Likewise, the vertical [4.1] vertical = ⅗ center-to-center … from [1.1] [4.2] vertical = ⅗ × 2𝒓 [4.3] vertical = 6⁄5𝒓 Armed with these, now the big triangle's base line consists of 3 parts [5.1] 28 = something + 8⁄5𝒓 + 𝒓 [5.2] 28 = something + 13⁄5𝒓 [5.3] something = 28 - 13⁄5𝒓 Connecting center of left circle to left corner of big triangle creates (with 𝒓 radius) 2 isosceles triangles; the point though is that [28 - 13⁄5𝒓] is the same for both the base-line bit and the hypotenuse line bit. So, working again with the hypotenuse line, the upper-right circle similarly creates an Isosceles triangle pair, whose edge side length … is reducible in algebra! [6.1] 𝒒 = 35 - (28 - 13⁄5𝒓) - 2𝒓 … reduces to [6.2] 𝒒 = 7 + ⅗𝒓 ALSO, note that [𝒒] is also defined from the '21' side of the big triangle as [7.1] 𝒒 = 21 - 𝒓 - 6⁄5𝒓 … from [4.3], reduces to [7.2] 𝒒 = 21 - 11⁄5𝒓 So, make these equal to each other, then find 𝒓! [8.1] 7 + ⅗𝒓 = 21 - 11⁄5𝒓 … rearrange [8.2] ((3 + 11)/5)𝒓 = (21 - 7) [8.3] (14⁄5)𝒓 = 14 … multiply by 5, divide by 14 [8.4] 𝒓 = 5 And THAT is the solution! No quadratics, no sines, no tangents, no nothing special. Just follow-the-parts algebra. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Let's find the radius R: . .. ... .... ..... First of all we apply the Pythagorean theorem to calculate the missing side length of the right triangle ABC: AC² = AB² + BC² = 28² + 21² = (4*7)² + (3*7)² = 4²*7² + 3²*7² = (3² + 4²)*7² = 5²*7² = (5*7)² = 35² ⇒ AC = 35 The area of the right triangle ABC can easily be calculated: A(ABC) = (1/2)*AB*BC = (1/2)*28*21 = 294 Now we divide the triangle ABC into a variety of figures: A(ABC) = A(ABO) + A(BCQ) + A(AEO) + A(CDQ) + A(DEOQ) + A(BOQ) Since AC is a tangent to both circles, we know that ∠AEO=∠CEO=∠ADQ=∠CDQ=90°. Because of EO=DQ=R and because the two circles have exact one point of intersection, we can conclude that DEOQ is a rectangle with the side lengths EO=DQ=R and DE=OQ=2*R. Therefore we obtain: A(ABO) = (1/2)*AB*OF = (1/2)*AB*R = (1/2)*28*R = 14*R A(BCQ) = (1/2)*BC*GQ = (1/2)*BC*R = (1/2)*21*R = 21*R/2 A(AEO) + A(CDQ) = (1/2)*AE*EO + (1/2)*CD*DQ = (1/2)*AE*R + (1/2)*CD*R = (1/2)*(AE + CD)*R = (1/2)*(AC − DE)*R = (1/2)*(AC − 2*R)*R = (1/2)*AC*R − R² = (1/2)*35*R − R² = 35*R/2 − R² A(DEOQ) = R*2*R = 2*R² For the calculation of A(BOQ) we need the height of the triangle ABC according to base AC: A(ABC) = (1/2)*AC*h(AC) ⇒ h(AC) = 2*A(ABC)/AC = 2*294/35 = 84/5 A(BOQ) = (1/2)*OQ*h(OQ) = (1/2)*(2*R)*(h(AC) − R) = R*(h(AC) − R) = R*(84/5 − R) = 84*R/5 − R² So finally we get: 294 = 14*R + 21*R/2 + 35*R/2 − R² + 2*R² + 84*R/5 − R² 294 = 294*R/5 ⇒ R = 5 Best regards from Germany
Exllent solution 👍. Thank you teacher 🙏.
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AC=√[21^2+28^2]=35
ED=2r
21 : 28 : 35 = 3 : 4 : 5 = 1.2r : 1.6r : 2r
GB=2.2r FB=2.6r
CD=CG=21-2.2r AE=AF=28-2.6r
21-2.2r+2r+28-2.6r=35
2.8r=14 r=5
Radius of the circle O , Q : 5
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nice solution my friend!
Very good... in just 7 lines!
Sin BAC = 3/5, cos BAC = 4/5
Draw lines from O to A and from Q to C. The lines bisects the angles A and C.
tan BAC/2 = sin / 1+cos = 3/5 / (1+ 4/5) = 1/3
tan ACB/2 = 4/5 / (1+ 3/5) = 1/2
AE = r/ tan BAC/2 -> AE = 3r
DC = r/ tan ACB/2 -> DC = 2r
AC = 3r + 2r +2r = 35 -> r = 5
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nice solution
Your video is very useful
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1/The triangle ABC is a 3x7-4x7-5x7 triple so, to make the calculation simpler:
Assume that r= (radius of the yellow circles/7) in a 3--4-5 triangle.
AO is the bisector of the angle CAB intersecting BC at point M. By using the bisector theorem:
CM/AC=BM/AB---> CM/5=BM/4= (CM+BM)/(5+4)=3/9= 1/3
----> BM=4/3
Consider the triangle AMB and AOF, they are similar so OF/BM= AF/AB= a/4---->
r/(4/3)= a/4----> a= 3r
2/ Similarly, the bisector CQ of the angle ACB intersects AB at point N, we have b=2r
3/AC= a+2r+b = 3r+2r+ 2r=5----> 7r=5---> r=5/7
The radius of the yellow circle= 7r= 5 units
The beauty of labeling everything! 🙂
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Nice challenge, thanks again!
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Excellent solution.
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Mind blowing 🥰😍,,MASTER💜
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Enjoyable...thanks...
Area ABO triangle: 28r/2
Area BCQ triangle: 21r/2
Area AOQC trapeze: (2r+35)r/2
Height OBQ triangle: 84/5-r so the area OBQ triangle: (84/5-r)2r/2
The sum of these areas: 21x28/2
So r=5
I've found the solution summing the areas inside ABC
Area(AOB)+area(CQB)+ Area trapezoid(ACQO) + Area(OQB) = Area(ABC)
28*r*1/2+21*r*1/2+(35+2r)*r*1/2+(84/5-r)*2r*1/2=21*28*1/2
14r+21/2r+35/2r+r^2+84/5r-r^2=294
R=5
The calculation would be a bit troublesome, GB=2r×(3/5)+r=11/5 r, FB=2r×(4/5)+r=13/5 r, so 21-11/5 r+28- 13/5 r+ 2r=35, 14=14/5 r, r=5.😊
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I think you made this a bit more complicated than necessary.
First, it helps to scale all the distances by 7 at the start, so the triangle sides are 3, 4 and 5.
Then at your 8:27, we have:
(4 - a - r)/2r = 4/5 (3 - b - r)/2r = 3/5
20 - 5a - 5r = 8r 15 - 5b - 5r = 6r
Add these two equations, giving 35 - 5(a + b) - 10r = 14r
Now from the length of the hypotenuse, 5 = a + b + 2r, i.e. a + b = 5 - 2r
35 - 5(5 - 2r) = 24r, 20 + 10r = 24r, 20 = 14r, r = 10/14 = 5/7.
Scaling back, r = 5.
Another sol:
Consider lime EO,OF,DQ,QG
Since they are radius they are equal hence the line is angle bisector using half angle sine formula we can easily see that the hypotenuse of triangle is 7r and is 35 hence r=5
Really interetsing and I am now grasping the idea of using equations to solve geometrical unknowns. Nobody ever explained it to me this way before. I am good at maths, but a sucession of bad maths teachers is cumulative, I'm afraid.
Radius (R) equal to 5 Linear Units.
First draw a Scalene Triangle [ABC] as here given, with sides 21 / 28 / 35, being the point A the Origin of the Coordinates (0 ; 0).
AC = 35, because sqrt(21^2 + 28^2) = sqrt(441 + 784) = sqrt(1.225) = 35. Check!!
Now, if I draw a a Line X = 7 I get a Square 21 by 21.
The Center (interception of the 2 Diagonals) of this Square have coordinates (16 ; 12).
Passing a Perpendicular Line to AC and in point (16 ; 12), the Center of the Square, we get a Triangle with Sides 15 / 20 / 25.
The Area of this Triangle is A = (Base * height) / 2 = (25 * 12) / 2 = 300 / 2 = 150 sq un
Now :
(15R + 20R + 25R) / 2 = 150
60R / 2 = 150
30R = 150
R = 150 / 30
R = 5
Answer:
The Radius is equal to 5 Linear Units.
As BC = 21 = 3(7), and AB = 28 = 4(7), ∆ABC is a 7:1 ratio Pythagorean triple triangle, and CA = 5(7) = 35. Draw radii OE, OF, QD, and QG. As radii, these segments are perpendicular to the tangents: AB to OF, BC to QG, and CA to QD and OE. Draw OQ. The quadrilateral OQDE is a rectangle, as OQ and DE are parallel and QD and OE are both perpendicular to DE. Thus OQ = DE = 2r.
Let M be a point on BC where OM is parallel to AB amd perpendicular to BC. Let N be a point on AB where QN is parallel to BC and perpendicular to AB. Let the intersection point of OM and QN be P.
Take note of trisngle ∆OPQ. All of its sides are parallel to the sides of ∆ABC, therefore all of its angles are the same, and the two triangles are similar. As OQ = 2r, OP = 4/5(2r) = 8r/5, and PQ = 3/5(2r) = 6r/5.
By two tangent theorem, GC and CD are congruent, and EA and AF are congruent. Let CD = GC = x and AF = EA = y. By observation, AB = y + 8r/5 + r, BC = r + 6r/5 + x, and CA = x + 2r + y. As we have numerical values for all three sides, we should be able to solve for r.
y + 8r/5 + r = 28 ---- (1)
y = 28 - 13r/5
r + 6r/5 + x = 21 ---- (2)
x = 21 - 11r/5
x + 2r + y = 35 ---- (3)
21 - 11r/5 + 2r + 28 - 13r/5 = 35
24r/5 - 2r = 49 - 35 = 14
14r/5 = 14
r = (5/14)14 = 5
AC = 35 (easy). Now, let's name t = angleBAC, we have tan(t) = 21/28 = 3/4, so tan(t/2) = 1/3 (easy to obtain with the formula giving tan(2.x))
So, a = AF = R/tan(t/2) = 3.R in triangle AFO.
Now let's name t' = angleACB, t' = 90° -t, so t'/2 = 45° -t/2 and tan(t') = 1/2 (easy to obtain with the formula giving tan(a -b))
So b = CG = R/tan(t') = 2.R in triangle CQG.
Now let'a use an orthonormam, center A, first axis (AB), second axis (AC). We have O(a;R) or O(3.R;R) and Q(28 -R; 21 -b) or Q(28 -R; 21 -2.R)
Then we have VectorOQ(28 -4.R; 21 -3.R) and OQ^2 = (28 -4.R)^2 + (21 -3.R)^2 = 784 -224.R +16.R^2 +441 -126.R +9.R^2 = 25.R^2 -350.R +1225
Or OQ = 2.R and OQ^2 = 4.R^2, then we have the equation: 21.R^2 -350.R +1225 = 0. Deltaprime = (-175)^2 -21.1225 = 4900 = (70)^2
So: R = (175 + 70)/21 = 245/21 = 35/3 which is rejected as in this case AF = a= 35 > AB =28, or R = (175 - 70)/21 = 5 which is our solution.
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great job, nice solution ... I really like it
I did the same as you to get AF = AE = 3R and CG = CD = 2R. But then I noticed that ED = 2R, so AC = 35 = 7R and R = 35/7 = 5 units 😉
Thank you!
Well-designed puzzle, let the right angled triangle be 3x-4x-5x, the answer r is integer, when x is multiple of 7, as r=5/7 x.🥰🥰🥰
AE=a,DC=b...(28-(a+r))^2+(21-(b+r))^2=(2r)^2....35=(a+r)+(b+r)...sostituisco risulta (28-(a+r))^2+((a+r)-14)^2=4r^2..(1eq)...2arctg(r/a)=arctg21/28...a=3r(2eq)...inserisco la 2)nella 1)...(28-4r)^2+(4r-14)^2=4r^2.svoogendo i calcoli rimane 32r^2-336r+980=4r^2....r^2-12r+35=0..r=6+1=7..r=6-1=5.…uhm..credo r=5, sia corretta
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Muy bueno profe.!
... Good day to you, Step(1): Triplet( 7*3 - 7*4 - 7*5 ) gives us I AC I = 35 u. ... I ED I = I OQ I = 2r ( r is radius circles ) ... I AE I = I AF I = X & I CD I = I CG I = Y ... so, I FB I = 28 - X & I GB I = 21 - Y ... Angle(BAC) = ARCTAN(3/4) ... Step(2): Creating inside right triangle(OQS) , where Angle(QSO) = 90 deg. ... Vertical distance V between Q and O ... I QS I = 21 - Y - r & Horizontal distance H between Q and O ... I OS I = 28 - X - r and Angle(SOQ) = Angle(BAC) = ARCTAN(3/4) ... last but not least I AC I = 35 = X + 2r + Y ... now TAN(SOQ) = TAN(BAC) = 3/4 = I QS I / I OS I ... 3/4 = ( 21 - Y - r ) / (28 - X - r ) ... after a few steps we obtain ... r = 3X - 4Y or 2r = 6X - 8Y & 2r = 35 - ( X + Y ) ... which gives us X - Y = 5 or Y = X - 5 .... so, r = 3X - 4(X - 5) = 20 - X .... finally ... TAN(OAF) = I OF I / I AF I = r / X = TAN(ARCTAN(3/4)/2) , so X = r / TAN(ARCTAN(3/4)/2) & recalling r = 20 - X ... after a few steps we obtain r = 20 / ( 1 + 1/TAN(ARCTAN(3/4)/2) ) = 20 / 4 = 5 u. ... r = 5 u. ... now it is time to watch your, I assume, much easier and clear explanation sir (lol), but reading other comments I guess I found the right answer for radius r in the end (lol) ... many thanks for your continuing interesting and fun geometry exercises ... best regards, Jan-W
Can you please prove the similarity
Naturally … yet another way, but essentially the SAME way as PreMath (but EASY! if you're good at manipulating basic algebra)
Draw line between centers of the circles.
[1.1] center-to-center length = 2𝒓
Figure length of hypotenuse of big triangle
[2.1] hypotenuse = √( 21² + 28² )
[2.2] hypotenuse = 35
This could also have been found by noting that 21 = 3×7 and 28 = 4×7, so the hypotenuse must be 5×7. Oh, well, I'm not intuitive enough to do it the 'easy way'.
Anyhow, now draw 2 more lines from the centers of the circles. A vertical one on the right circle and a horizontal one from the left circle. They intersect, forming a similar triangle to the big triangle. Similar means 'proportionate to'. So…
[3.1] horizontal = ⅘ center-to-center … from [1.1]
[3.2] horizontal = ⅘ × 2𝒓
[3.3] horizontal = 8⁄5𝒓
Likewise, the vertical
[4.1] vertical = ⅗ center-to-center … from [1.1]
[4.2] vertical = ⅗ × 2𝒓
[4.3] vertical = 6⁄5𝒓
Armed with these, now the big triangle's base line consists of 3 parts
[5.1] 28 = something + 8⁄5𝒓 + 𝒓
[5.2] 28 = something + 13⁄5𝒓
[5.3] something = 28 - 13⁄5𝒓
Connecting center of left circle to left corner of big triangle creates (with 𝒓 radius) 2 isosceles triangles; the point though is that [28 - 13⁄5𝒓] is the same for both the base-line bit and the hypotenuse line bit.
So, working again with the hypotenuse line, the upper-right circle similarly creates an Isosceles triangle pair, whose edge side length … is reducible in algebra!
[6.1] 𝒒 = 35 - (28 - 13⁄5𝒓) - 2𝒓 … reduces to
[6.2] 𝒒 = 7 + ⅗𝒓
ALSO, note that [𝒒] is also defined from the '21' side of the big triangle as
[7.1] 𝒒 = 21 - 𝒓 - 6⁄5𝒓 … from [4.3], reduces to
[7.2] 𝒒 = 21 - 11⁄5𝒓
So, make these equal to each other, then find 𝒓!
[8.1] 7 + ⅗𝒓 = 21 - 11⁄5𝒓 … rearrange
[8.2] ((3 + 11)/5)𝒓 = (21 - 7)
[8.3] (14⁄5)𝒓 = 14 … multiply by 5, divide by 14
[8.4] 𝒓 = 5
And THAT is the solution! No quadratics, no sines, no tangents, no nothing special. Just follow-the-parts algebra.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Let's find the radius R:
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First of all we apply the Pythagorean theorem to calculate the missing side length of the right triangle ABC:
AC² = AB² + BC² = 28² + 21² = (4*7)² + (3*7)² = 4²*7² + 3²*7² = (3² + 4²)*7² = 5²*7² = (5*7)² = 35²
⇒ AC = 35
The area of the right triangle ABC can easily be calculated:
A(ABC) = (1/2)*AB*BC = (1/2)*28*21 = 294
Now we divide the triangle ABC into a variety of figures:
A(ABC) = A(ABO) + A(BCQ) + A(AEO) + A(CDQ) + A(DEOQ) + A(BOQ)
Since AC is a tangent to both circles, we know that ∠AEO=∠CEO=∠ADQ=∠CDQ=90°. Because of EO=DQ=R and because the two circles have exact one point of intersection, we can conclude that DEOQ is a rectangle with the side lengths EO=DQ=R and DE=OQ=2*R. Therefore we obtain:
A(ABO)
= (1/2)*AB*OF
= (1/2)*AB*R
= (1/2)*28*R
= 14*R
A(BCQ)
= (1/2)*BC*GQ
= (1/2)*BC*R
= (1/2)*21*R
= 21*R/2
A(AEO) + A(CDQ)
= (1/2)*AE*EO + (1/2)*CD*DQ
= (1/2)*AE*R + (1/2)*CD*R
= (1/2)*(AE + CD)*R
= (1/2)*(AC − DE)*R
= (1/2)*(AC − 2*R)*R
= (1/2)*AC*R − R²
= (1/2)*35*R − R²
= 35*R/2 − R²
A(DEOQ) = R*2*R = 2*R²
For the calculation of A(BOQ) we need the height of the triangle ABC according to base AC:
A(ABC) = (1/2)*AC*h(AC)
⇒ h(AC) = 2*A(ABC)/AC = 2*294/35 = 84/5
A(BOQ)
= (1/2)*OQ*h(OQ)
= (1/2)*(2*R)*(h(AC) − R)
= R*(h(AC) − R)
= R*(84/5 − R)
= 84*R/5 − R²
So finally we get:
294 = 14*R + 21*R/2 + 35*R/2 − R² + 2*R² + 84*R/5 − R²
294 = 294*R/5
⇒ R = 5
Best regards from Germany
I used trigonometry to solve. Very good see other solution! Thanks professor!!
⭐️
As an old framing carpenter I can relate to 3-4-5 rule. Used it 1,000’s of times.
Excellent!
Thanks ❤️
👇1 like =10 marks increase in exams .....do not take risk as exams are important
Excellent!
Thanks ❤️
Is this like an electronic Apple for the teacher?
Bet you sat in the front of the class… 😁
yes
Now I wonder: Could you solve for r for any number of circles (as long as they were tightly inscribed)?
isn't it 49-24r/5=35?
I'll say r=5.
In Hindi language plz