36/5=7.2, by considering the triangle with sides 6, 6, x with angle 2t, where tan t=1/2, so cos 2t=3/5, by cosine rule 6^2+x^2-2 6 x 3/5=6^2, then x=36/5.😊
Other possible non trigonometric solution: Triangles BOP and BPC are congruent having BP in common, OP and PC are radii of semicircle, BO = BC for two tangent theorem, so BP is the bisector of B angle. Applying the bisector theorem we get: OB/BE = OP/PE 6/x = 3/y x/y = 6/3 = 2 then PE = y and BE = 2*PE = 2y Applying the Pythagorean theorem on BOE we can write: 6² + (3+y)² = (2y)² y² -2y - 15 = = y = 5 BO = 6, OE = 3+5 = 8, BE = 2*5 = 10 now trace the radius of the quarter circle perpendicular to chord BD, and call it OH triangle BOH is similar to BOE having B angle in common and being right, so: OB : BE = BH : OB 6 : 10 = BH : 6 BH = X/2 = 18/5 then 2X = 2*18/5 = 36/5
Another solution: Connect OC and extend it, intersecting the big circle at E(the one close to C) and F. connect CA. Triangle OCA is similar to BOP. So, OC =2*CA, OC^2 + CA^2 = OA^2, OC = 6/sqrt(5). According to Intersecting Chords Theorem， CF*CE= BC*CD, (6 + 6/sqrt(5)) * (6 - 6/sqrt(5)) = 6 * CD, CD = 1.2, BD = 7.2
36/5 , considering the triangle BOP with sides 6,3,3√ 5, we can find cos alpha = 2√ 5/5, then cos 2alpha = 3/5 (being BOP and BPC congruent) and then X = 2*6*cos 2alpha = 12*3/5 = 36/5 (tracing the perpendicular from O to line BD)
Another solution: Extend BO to Q so that BQ is diameter of the large circle. Connect DQ. Extend CP, intersecting BQ at R. Angle BCR = BDQ = 90, CR || DQ. Triangle ROP and RCB are similar, we can get RO = 4, BR:BQ = BC:BD, 10:12=6:BD, BD = 7.2
This one isn’t too bad to bash out algebraically. Placing the origin at O and the coordinate axes along the two given radii of the large circle, we can find that an equation of the smaller circle is s = x^2+y^2-6x=0. Using Joachimsthal’s notation, the tangents to this circle from B=(0,6) are given by the equation (s_B)^2=s·s_BB, i.e., (0x+6y-3(x+0))^2 = (x^2+y^2-6x)(0^2+6^2-6·0), or (6y-3x)^2 = 36(x^2+y^2-6x). Rearranging and factoring produces -9x(3x+4y-24)=0. From there, computing the coordinates of D is a simple matter of finding the intersection of 3x+4y-24=0 with the large circle. Not the most elegant solution, but straightforward.
1/ Just label the angle PBC=alpha so the angle OBC =2 alpha --------> tan aplha =1/2 -----> tan 2aplpha= 1/(1-1/4) =.4/3 2/ Drop OH perpendicular to chord AD we have BH=BD/2-----> OH/BH= 4/3------> OH= 4BH/3 -----> sq OH + sq BH= sq (4BH/3) + sqBH =sqOB-------> 25 sqBH =9x36 -----> BH=18/5 BD=36/5= 7.2 cm
Wow! That was a neat way to do it. My approach was to draw BP to find cos(OBP). Use this and the double angle cosine formula to find cos(OBC). Then reflect the quarter circle over OA. Now draw a segment from D to the other end of the diameter just formed. This makes a right angle at point D so that BD = (2*6)cos(OBC) = 36/5.
36/5=7.2, by considering the triangle with sides 6, 6, x with angle 2t, where tan t=1/2, so cos 2t=3/5, by cosine rule 6^2+x^2-2 6 x 3/5=6^2, then x=36/5.😊
Geometry is so much fun . . . with so many alternate solutions.
Thanks for sharing it.
Other possible non trigonometric solution:
Triangles BOP and BPC are congruent having BP in common, OP and PC are radii of semicircle, BO = BC for two tangent theorem, so BP is the bisector of B angle. Applying the bisector theorem we get:
OB/BE = OP/PE
6/x = 3/y
x/y = 6/3 = 2 then PE = y and BE = 2*PE = 2y
Applying the Pythagorean theorem on BOE we can write:
6² + (3+y)² = (2y)²
y² -2y - 15 = =
y = 5
BO = 6, OE = 3+5 = 8, BE = 2*5 = 10
now trace the radius of the quarter circle perpendicular to chord BD, and call it OH
triangle BOH is similar to BOE having B angle in common and being right, so:
OB : BE = BH : OB
6 : 10 = BH : 6
BH = X/2 = 18/5
then 2X = 2*18/5 = 36/5
Geometry has so many alternate solution. Its so much fun checking them.
Thanks for sharing it.
Another solution: Connect OC and extend it, intersecting the big circle at E(the one close to C) and F. connect CA. Triangle OCA is similar to BOP. So, OC =2*CA, OC^2 + CA^2 = OA^2, OC = 6/sqrt(5). According to Intersecting Chords Theorem， CF*CE= BC*CD, (6 + 6/sqrt(5)) * (6 - 6/sqrt(5)) = 6 * CD, CD = 1.2, BD = 7.2
Geometry has s many alternate solution. Its fun checking them.
Thanks for sharing it.
36/5 , considering the triangle BOP with sides 6,3,3√ 5, we can find cos alpha = 2√ 5/5, then cos 2alpha = 3/5 (being BOP and BPC congruent) and then X = 2*6*cos 2alpha = 12*3/5 = 36/5 (tracing the perpendicular from O to line BD)
Geometry is so much fun . . . with so many alternate solutions.
Thanks for sharing it.
Very interesting problem and nice solution.
Glad you like it
Another solution: Extend BO to Q so that BQ is diameter of the large circle. Connect DQ. Extend CP, intersecting BQ at R. Angle BCR = BDQ = 90, CR || DQ. Triangle ROP and RCB are similar, we can get RO = 4, BR:BQ = BC:BD, 10:12=6:BD, BD = 7.2
Woow . . . There is so much fun in checking the alternate solutions.
Thanks for sharing it.
Brilliant !!
Glad that you liked it.
This one isn’t too bad to bash out algebraically. Placing the origin at O and the coordinate axes along the two given radii of the large circle, we can find that an equation of the smaller circle is s = x^2+y^2-6x=0. Using Joachimsthal’s notation, the tangents to this circle from B=(0,6) are given by the equation (s_B)^2=s·s_BB, i.e., (0x+6y-3(x+0))^2 = (x^2+y^2-6x)(0^2+6^2-6·0), or (6y-3x)^2 = 36(x^2+y^2-6x). Rearranging and factoring produces -9x(3x+4y-24)=0. From there, computing the coordinates of D is a simple matter of finding the intersection of 3x+4y-24=0 with the large circle. Not the most elegant solution, but straightforward.
Geometry has so many alternate solution.
Its so much fun checking them.
Thanks for sharing it.
1/ Just label the angle PBC=alpha so the angle OBC =2 alpha --------> tan aplha =1/2 -----> tan 2aplpha= 1/(1-1/4) =.4/3
2/ Drop OH perpendicular to chord AD we have BH=BD/2-----> OH/BH= 4/3------> OH= 4BH/3
-----> sq OH + sq BH= sq (4BH/3) + sqBH =sqOB-------> 25 sqBH =9x36 -----> BH=18/5
BD=36/5= 7.2 cm
Woow . . . great
Thanks for sharing it.
Wow! That was a neat way to do it. My approach was to draw BP to find cos(OBP). Use this and the double angle cosine formula to find cos(OBC). Then reflect the quarter circle over OA. Now draw a segment from D to the other end of the diameter just formed. This makes a right angle at point D so that BD = (2*6)cos(OBC) = 36/5.
Woow . . . There is so much fun in checking the alternate solutions.
Thanks for sharing it.
tan α/2 = r/R = 3/6 = 1/2
α = 53,13°
Triangle OBD is isósceles (R,R,x),
Cosine rule :
x² = 2R² [1 - cos (180-2α)]
x² = 2 6² (1-cos 73,74°)
x = 7,2 cm ( Solved √ )
Good one.
tan α/2 = r/R = 3/6 = 1/2
α = 53,13°
cos α = x / 2R
x = 2R cos α
x = 12. cos 53,13°
x = 7,2 cm ( Solved √ )
I hadn't watched the video yet and came up with 7.2
Great . . . Nothing can beat the fun in solving math puzzle and satisfaction in tallying the answer.
That was fun !
Glad that you liked it.
I loved the thumbnail.
Glad that you like it.