Solution by half interval search: At 2:20 in the video, a perpendicular has been dropped to PQ, dividing it into 2 equal length segments, PN and NQ, each length 3. Drop a perpendicular to BQ and label the intersection as point R. BQ has been divided into 2 equal length segments, BR and QR, each length 5. Construct OQ. ΔBOR and ΔQOR are congruent by side-angle-side. From congruency,
Let C be the point diagonally opposite to point B, and let H be the perpendicular projection of point Q onto BC, and let r be the radius of the semicircle. Since the trapezoid BQPC is isosceles because PQ is parallel to BC, then BH=CA=(2r-6)/2=r-3, and we have in the right triangle BQC BH*BC=BQ², and from it (r-3)*2r=100, so r²-3r-50=0 so r=(3+√209)/2 .
Nice One : ) While using symmetry I set AP = "h" and solved h²= √(100 - h²)(6 + √(100 - h²). From this I was able to find h and then uncover that √(h² + 9) = (3 + √209)/2.
Thank you for this problem and your well-presented solution. required radius = r 3.3 = r . r - AP.AP because MO = AO where M is the midpoint of PQ, and O is the centre of the circle (of which we are given the top half.) and because PQ and the diameter which goes through M intersect at M. (r -3 )(r+3)= AP.AP gives no new information but looks at A as the intersection of AP and B'A OB, the diameter through A B'P is a mirror image of BP. In triangle B'PA PA.PA + B'A.B'A = B'P.B'P =10.10=100 B'A.B'A = 100 - AP.AP =100+9 - r.r = 109 -r.r and r - 3 = B'A so 109 - r.r = r.r - 6 r +9 2r.r - 6r -100 =0 r.r - 3r - 50 =0 r = (1/2)(3 +/- sqrt (9 +200) r > 0 so the radius , r = ( sqrt (209) + 3 )/2
Let C be the center of the semicircle diameter. Drop a perpendicular QD onto AB. As |CP| = |CQ| = r, the midpoint M of PQ is on a line CM parallel to AP and QD and therefore |AC| = |CD| = 3. Let E be the point on the left end of the diameter and consider triangle EBQ. EBQ is a right trinagle with height h = QD and we have |ED| = r + 3, |DB| = r - 3. So h² = (r - 3)(r + 3) = r² - 9. Considering right triangles EDQ and EBQ, we can finally establsih that (2r)² - 10² = (r + 3)² + r² - 9. This simplifies to the quadratic equation r² - 3r -50 = 0 with the only sensible solution r = (3 + √209) / 2 as the second solution is negative.
Solution : Draw a Vertical Line passing by Point Q, parallel to AP. Call the the intercepting Point on Line AB, Point A'. Call the Center of the Semicircle C. AC = CA' = PQ / 2 = 3 AP = h CQ = R(adius) h^2 = R^2 - 9 h^2 = 100 - (R - 3)^2 R^2 - 9 = 100 - R^2 + 6R - 9 2R^2 = 100 + 6R 2R^2 - 6R - 100 = 0 R^2 - 3R - 50 = 0 Positive Solution : R = (3 + sqrt(209)) / 2 R ~ 8,7284
It's funny how a successful solutions of these problems depends on picking up the small facts. I missed the fact that OQ is a radius, and ended up with an unwieldy 4th order equation.
Solution by half interval search: At 2:20 in the video, a perpendicular has been dropped to PQ, dividing it into 2 equal length segments, PN and NQ, each length 3. Drop a perpendicular to BQ and label the intersection as point R. BQ has been divided into 2 equal length segments, BR and QR, each length 5. Construct OQ. ΔBOR and ΔQOR are congruent by side-angle-side. From congruency,
Pytagorean theorem, twice:
R² - (6/2)² = 10² - (R - 6/2)²
R² - 3² = 10² - (R² - 6R + 3²)
2R² - 6R -10² = 0
R² - 3R -50 = 0 ---> R = 8,7284 cm
Let C be the point diagonally opposite to point B, and let H be the perpendicular projection of point Q onto BC, and let r be the radius of the semicircle. Since the trapezoid BQPC is isosceles because PQ is parallel to BC, then BH=CA=(2r-6)/2=r-3, and we have in the right triangle BQC BH*BC=BQ², and from it (r-3)*2r=100, so r²-3r-50=0 so r=(3+√209)/2 .
Nice One : )
While using symmetry I set AP = "h" and solved h²= √(100 - h²)(6 + √(100 - h²). From this I was able to find h and then uncover that √(h² + 9) = (3 + √209)/2.
2*arcsin(10/(2*x))+arcsin(6/(2*x))=Pi/2
2*arcsin(5/x)+arcsin(3/x)=Pi/2
2*arcsin(5/x)=Pi/2-arcsin(3/x)
cos(2*arcsin(5/x))=cos(Pi/2-arcsin(3/x))
1-2*(sin(arcsin(5/x)))^2=sin(arcsin(3/x))
1-2*(5/x)^2=3/x
1-2*25/x^2=3/x
1-50/x^2=3/x
x^2-3*x-50=0 (same equation than video at 8:12)
answer : x=(3+sqrt(209))/2
Thank you for this problem and your well-presented solution.
required radius = r
3.3 = r . r - AP.AP because MO = AO where M is the midpoint of PQ, and O is the centre of the circle (of which we are given the top half.)
and because PQ and the diameter which goes through M intersect at M.
(r -3 )(r+3)= AP.AP gives no new information but looks at A as the intersection of AP and B'A OB, the diameter through A
B'P is a mirror image of BP. In triangle B'PA PA.PA + B'A.B'A = B'P.B'P =10.10=100
B'A.B'A = 100 - AP.AP =100+9 - r.r = 109 -r.r and r - 3 = B'A
so 109 - r.r = r.r - 6 r +9 2r.r - 6r -100 =0 r.r - 3r - 50 =0 r = (1/2)(3 +/- sqrt (9 +200)
r > 0 so the radius , r = ( sqrt (209) + 3 )/2
Your thinking is very creative. I enjoyed reading your logical steps : )
@@oscarcastaneda5310 : ) Thank you for telling me. I enjoy typing and thinking together.
Let C be the center of the semicircle diameter. Drop a perpendicular QD onto AB. As |CP| = |CQ| = r, the midpoint M of PQ is on a line CM parallel to AP and QD and therefore |AC| = |CD| = 3. Let E be the point on the left end of the diameter and consider triangle EBQ. EBQ is a right trinagle with height h = QD and we have |ED| = r + 3, |DB| = r - 3. So h² = (r - 3)(r + 3) = r² - 9. Considering right triangles EDQ and EBQ, we can finally establsih that (2r)² - 10² = (r + 3)² + r² - 9. This simplifies to the quadratic equation r² - 3r -50 = 0 with the only sensible solution r = (3 + √209) / 2 as the second solution is negative.
A very concise and logical explanation : )
円の中心をO、QからABに垂線を引いて交点をDとしAB₌a、AP₌b、半径rとする。
(a₋r)^2₊b^2₌r^2 a^2₊b^2₋2ar₊r^2₌r^2よりa^2₊b^2₌2ar ①
(a₋6)^2₊b^2₌100 a^2₊b^2₋12ar₊36₌100よりa^2₊b^2₌64₊12ar ②
①、②より2ar₌64₊12ar ar₌32₊6ar a₌32/(r₋6) ③
円の左端の点をEとすると△BQE∽△BQDよりa₋6：10₌10：2r
100₌2r(a₋6) a₌(6r₊50)/r ④
③、④より(6r₊50)/r₌32/(r₋6) 6r^2₋36r₊50r₋300₌32r 6(r^2₋3r₋50)₌0
r₌(3₊√209)/2
Solution :
Draw a Vertical Line passing by Point Q, parallel to AP. Call the the intercepting Point on Line AB, Point A'.
Call the Center of the Semicircle C.
AC = CA' = PQ / 2 = 3
AP = h
CQ = R(adius)
h^2 = R^2 - 9
h^2 = 100 - (R - 3)^2
R^2 - 9 = 100 - R^2 + 6R - 9
2R^2 = 100 + 6R
2R^2 - 6R - 100 = 0
R^2 - 3R - 50 = 0
Positive Solution :
R = (3 + sqrt(209)) / 2
R ~ 8,7284
It's funny how a successful solutions of these problems depends on picking up the small facts. I missed the fact that OQ is a radius, and ended up with an unwieldy 4th order equation.
2r/10= 10/(r-3).
PA²=r²-3²=10²-(r-3)²---> r=(3+√209)/2.
Gracias y un saludo cordial.
(6)^2 (10)^2={36+100}=136ABPQ/90=1.46ABPQ 1.2^23 1.2^23^1 1.2^1^1 2^1 (ABPQ ➖ 2ABPQ+1).
R^2-3^2=10^2-(R-3)^2……R=(3+√209)/2
1.st Euclid theorem 100 =x(2x+6) 2x2 +6x -100 = 0 x = (-3+√209 )/2 r = (√209 + 3)/2