At 15:20, Math Booster has 2 solutions and discards the larger solution, because point P must be outside square ABCD. Actually, the second solution is for the case where P lies inside the square and is the given distances away from points A, B, and C.
This definitely a problem that shows a subtle combination of algebra and geometric reasoning. Impossible means that it is easier than it looks. And lots of practice required in order to just comprehend how and why it is easier than it looks.
May an analytic geometry approach. assume coordinates as P(x, y), B(0,0), A(-h,0), and C(0, -h) . three unknown numbers x/y/h in three equations of the Pythagorean theorem. similar solution but easier to think about.
Thank you for the nice problem and its solution. My solution is different, thus maybe worth sharing here: I tried to exploit the observation that the angles BAP, BCP and APC add up to 90° by rotating the triangle BCP by 90° clockwise around B, i.e. such that C moves to A and P to another point, call it P'. One then has the triangle APP', whose sides are all known as the triangle PBP' is rectangular and isosceles, meaning that |PP'| = 3*sqrt(2) and whose angle PAP' is equal to the sum of the angles BCP and BAP. Applying the cosine rule, one has for triangle PAP' that (3*sqrt(2))^2 = 5^2 + 6^2 - 2*5*6*cos([angle PAP']), which means that cos([angle PAP']) = 43/60. Now one can come back to triangle APC and apply the cosine rule again, meaning that |AC|^2 = 5^2 + 6^2 - 2*5*6*cos([angle APC]) = 5^2 + 6^2 - 2*5*6*sin([angle BAP]+[angle BCP]) = 61 - sqrt(17*103), the second equation being the expression of the observation that the angles BAP, BCP and APC add up to 90° and the third equation resulting from sin([angle PAP']) = sqrt(1 - (cos([angle PAP']))^2) = sqrt(17*103)/60. The area of the square ADCB is then equal to |AC|^2/2 = (61 - sqrt(17*103))/2.
instigators trailer? i have written 4 equations with 4 unknown numbers and assumed that the common intersection point "P" is at 0;0: 10 print "math booster-impossible geometry problem-square given by 3 distances from 1 point" 20 l1=5:l2=3:l3=6:sw=l1^2/(l1+l2+l3)/10:xm1=sw:dim x(3),y(3):goto 60 30 ym1=sqr(l1^2-xm1^2):p=-(xm1+ym1):q=((xm1+ym1)^2-l3^2)/2 40 ym2=-p/2+sqr(abs(p*p/4-q)):xm2=xm1-ym2+ym1 50 dgu1=(xm2/l2)^2:dgu2=(ym1/l2)^2:dg=dgu1+dgu2-1:return 60 gosub 30 70 dg1=dg:xm11=xm1:xm1=xm1+sw:xm12=xm1:gosub 30:if dg1*dg>0 then 70 80 xm1=(xm11+xm12)/2:gosub 30:if dg1*dg>0 then xm11=xm1 else xm12=xm1 90 if abs(dg)>1E-10 then 80 100 print xm1;"%";ym1;"%";xm2;"%";ym2:la=xm1-xm2:print "laenge l=";la 110 x(0)=xm1:y(0)=ym1:x(1)=xm2:y(1)=ym1:x(2)=xm2:y(2)=ym2:x(3)=xm1:y(3)=ym2 120 xmin=(l1+l2+l3)/3:ymin=xmin:xmax=xmin:ymax=ymin 130 for a=0 to 3:if x(a)xmax then xmax=x(a) 150 if y(a)ymax then ymax=y(a) 170 next a:if xmin>0 then xmin=0 180 if ymin>0 then ymin=0 190 if xmax=xmin or ymax=ymin then else 230 200 if xmax=xmin then else 230 210 mass=850/(ymax-ymin):goto 240 220 mass=1200/(xmax-xmin):goto 240 230 masx=1200/(xmax-xmin):masy=850/(ymax-ymin):if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
You didn't even watch the full video to see if it's nice or not, it's literally uploaded 3 minutes ago and you commented 1 min ago and possibly watched for 2 mins
At 15:20, Math Booster has 2 solutions and discards the larger solution, because point P must be outside square ABCD. Actually, the second solution is for the case where P lies inside the square and is the given distances away from points A, B, and C.
Let _θ = ∠ABP_
Using the cosine rule:
In _ΔABP: 5² = x² + 3² - 6xcos(θ)_
⇒ _6xcos(θ) = x² - 16_
S.B.S : _36x²cos²(θ) = x⁴ - 32x² + 256_ ... ①
In _ΔBCP: 6² = x² + 3² - 6xcos(270° - θ)_
⇒ _6xsin(θ) = -x² + 27_
S.B.S : _36x²sin²(θ) = x⁴ - 54x² + 729_ ... ②
Adding ① and ②:
_36x² = 2x⁴ - 86x² + 985_
⇒ _2x⁴ - 122x² + 985 = 0_
⇒ *area of **_ABCD = x² = ½(61 ± √1751)_*
s = side length of square. Using coordinates we get:
A = (0,s), B = (s,s), C = (s,0), D = (0,0)
P is point of intersection of circles centered at A, B, C with radii 5, 3, 6 respectively. These circles have equations:
(1) x² + (y−s)² = 25
(2) (x−s)² + (y−s)² = 9
(3) (x−s)² + y² = 36
(1) − (2) gives
x² − (x² − 2xs + s²) = 25 − 9 → x = (16+s²)/(2s)
(3) − (2) gives
y² − (y² − 2ys + s²) = 36 − 9 → y = (27+s²)/(2s)
Plugging values of x and y into (2) we get
((16+s²)/(2s) − s)² + ((27+s²)/(2s) − s)² = 9
((16−s²)/(2s))² + ((27−s²)/(2s))² = 9
(256−32s²+s⁴)/(4s²) + (729−54s²+s⁴)/(4s²) = 9
2s⁴ − 86s² + 985 = 36s²
2s⁴ − 122s² + 985 = 0
s² = (61+√1751)/2 ≈ 51.42 or
s² = (61−√1751)/2 ≈ 9.58
Since s < 5, then s² < 25. Therefore:
Area(ABCD) = s² = *(61−√1751)/2 ≈ 9.58*
Note that sin
This definitely a problem that shows a subtle combination of algebra and geometric reasoning. Impossible means that it is easier than it looks. And lots of practice required in order to just comprehend how and why it is easier than it looks.
Sir I have a math problem But I am not able to solve it can you solve it but how I can send the problem to you?
May an analytic geometry approach. assume coordinates as P(x, y), B(0,0), A(-h,0), and C(0, -h) . three unknown numbers x/y/h in three equations of the Pythagorean theorem. similar solution but easier to think about.
Thank you for the nice problem and its solution. My solution is different, thus maybe worth sharing here: I tried to exploit the observation that the angles BAP, BCP and APC add up to 90° by rotating the triangle BCP by 90° clockwise around B, i.e. such that C moves to A and P to another point, call it P'. One then has the triangle APP', whose sides are all known as the triangle PBP' is rectangular and isosceles, meaning that |PP'| = 3*sqrt(2) and whose angle PAP' is equal to the sum of the angles BCP and BAP. Applying the cosine rule, one has for triangle PAP' that (3*sqrt(2))^2 = 5^2 + 6^2 - 2*5*6*cos([angle PAP']), which means that cos([angle PAP']) = 43/60. Now one can come back to triangle APC and apply the cosine rule again, meaning that |AC|^2 = 5^2 + 6^2 - 2*5*6*cos([angle APC]) = 5^2 + 6^2 - 2*5*6*sin([angle BAP]+[angle BCP]) = 61 - sqrt(17*103), the second equation being the expression of the observation that the angles BAP, BCP and APC add up to 90° and the third equation resulting from sin([angle PAP']) = sqrt(1 - (cos([angle PAP']))^2) = sqrt(17*103)/60. The area of the square ADCB is then equal to |AC|^2/2 = (61 - sqrt(17*103))/2.
Uma questão bem desafiante! Parabéns!!!
instigators trailer?
i have written 4 equations with 4 unknown numbers and assumed
that the common intersection point "P" is at 0;0:
10 print "math booster-impossible geometry problem-square given by 3 distances from 1 point"
20 l1=5:l2=3:l3=6:sw=l1^2/(l1+l2+l3)/10:xm1=sw:dim x(3),y(3):goto 60
30 ym1=sqr(l1^2-xm1^2):p=-(xm1+ym1):q=((xm1+ym1)^2-l3^2)/2
40 ym2=-p/2+sqr(abs(p*p/4-q)):xm2=xm1-ym2+ym1
50 dgu1=(xm2/l2)^2:dgu2=(ym1/l2)^2:dg=dgu1+dgu2-1:return
60 gosub 30
70 dg1=dg:xm11=xm1:xm1=xm1+sw:xm12=xm1:gosub 30:if dg1*dg>0 then 70
80 xm1=(xm11+xm12)/2:gosub 30:if dg1*dg>0 then xm11=xm1 else xm12=xm1
90 if abs(dg)>1E-10 then 80
100 print xm1;"%";ym1;"%";xm2;"%";ym2:la=xm1-xm2:print "laenge l=";la
110 x(0)=xm1:y(0)=ym1:x(1)=xm2:y(1)=ym1:x(2)=xm2:y(2)=ym2:x(3)=xm1:y(3)=ym2
120 xmin=(l1+l2+l3)/3:ymin=xmin:xmax=xmin:ymax=ymin
130 for a=0 to 3:if x(a)xmax then xmax=x(a)
150 if y(a)ymax then ymax=y(a)
170 next a:if xmin>0 then xmin=0
180 if ymin>0 then ymin=0
190 if xmax=xmin or ymax=ymin then else 230
200 if xmax=xmin then else 230
210 mass=850/(ymax-ymin):goto 240
220 mass=1200/(xmax-xmin):goto 240
230 masx=1200/(xmax-xmin):masy=850/(ymax-ymin):if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
Thank for watching
Beautiful question!!!
Nice video keep it up
You didn't even watch the full video to see if it's nice or not, it's literally uploaded 3 minutes ago and you commented 1 min ago and possibly watched for 2 mins
@@Arandomguy1ylI have done the answer in my head and that's why I watched it by speeding up by a factor of 10seconds
AREA SQUARE ABCD=(61-√1751)/2
A≈9.578 (51.422) !!! 😁
Note that cos
@@jimleahy3858🤔???
If it's impossible then how you expect us to solve it. Are you from a mental home