Easily prove that triangle ABQ is isosceles. Let AQ=QB=QC=x. The area of triangle ABC is AC*BP/2=20. Therefore, AC*BP=40. Therefore, BP=20/x. Applying the Pythagorean theorem to triangle BPQ, we find x²=(x-2)²+(20/x)². This is equivalent to x³-x²-100=0. That is, (x-5)(x²+4x+20)=0. Therefore, x=5. Since PQ=3 and BP=4, the area of triangle BPQ is (3*4)/2=6.
Small shortcut: Q is the center of the circle passing thru A, B and C and AC is the diameter. So c = a + 2. I appreciate for your daily videos. Thank you for your time.
@13:36 that c=2+a is easy to see from the equality QC and QB. If we drop a perpendicular on BC we will have OM as a middle line of a triangle which is equal to a half of a base AB.
At about 7:38, Math Booster makes an educated guess that b = 4 is a root. The root can be found without guessing by doing a half interval search. Start the search with 2 values of b, one where b³ + 4b - 80 is negative and one where it is positive. Since b³ + 4b - 80 is continuous, there must be a zero crossing, a value of b where b³ + 4b - 80 is zero, somewhere between our two values. We know that as b increases in absolute value, b³ will dominate, so there must be negative values of b which produce a negative value of b³ + 4b - 80 and positive values of b which produce a positive value of b³ + 4b - 80. We try b = -10 and find b³ + 4b - 80 = -1040 and, for b = 10, b³ + 4b - 80 = 960. So, try -10 and 10 as limits. We start by testing b = 0 (half way between -10 and +10). If positive, we'll go half way to -10 to try our next number. If negative, we go half way to 10. We cut the increment to half. So, after finding that b = 0 produces a negative value, we'll try b = 5. The next try will be either 5 - 2.5 or 5 + 2.5. So, here is a computer program written in python to find b where b³ + 4b - 80 is approximately 0: # search range is b = -10 to +10 # start at the middle, b = 0 b = 0.0 db = 5.0 # define function to compute delta = ∆ def compute_delta(b): # ∆ = b³ + 4b - 80, rewrite in python notation delta = b**3 + 4*b -80 return(delta) for n in range(0,51): # after 50 divisions by 2, db is sufficiently small delta = compute_delta(b) if (delta > 0): b = b - db else: b = b + db db = db/2 b = round(b , 15) # round b to 15 decimal places formatted = "{:.15f}".format(b) print(f"b = ",formatted) Output: b = 3.999999999999999 === Code Execution Successful === So, we round up to 4, try b = 4 and find that b³ + 4b - 80 is exactly 0. We can factor out (b - 4) to get a quadratic, which we know how to solve. In this case, the roots of the quadratic are not real numbers, so b = 4 is the only real root. One viewer came up with a solution that required finding roots for x³ -x² -100. The python program can be rewritten to solve x³ -x² -100 = 0 for x by replacing x with b and changing one line. The line delta = b**3 + 4*b -80 becomes delta = b**3 - b**2 - 100: Output b = 5.000000000000004 === Code Execution Successful === Round down to 5 and try b = 5. An afterthought, refining the limits: we know that b or x must be positive, so we could try a lower limit of 0, which produces a negative value for our third order expression. Choose an upper value of b or x which produces a positive value.
Since we have a right triangle and QC = QB then QC = QA. Let's Call each of these "x". Area of △ABQ = 10 so hx = 20. Also 2(2x - 2) = h², and since hx = 20, h²x² = 400. So 2x²(2x - 2) = 400 and x²(x - 1) = 100 from which x = 5 so h = 4. With this A = h(x - 2)/2 = 6.
Let u = BC, then BA = (2.area of ABC)/u = 40/u. Triangles APB and ABC are similar, so AP/BP = AB/BC = (u^2)/40, so 2/BP = 40/(u^2) we then have that BP = (u^2)/20. Now in triangle APB: AB^2 = AP^2 + BP^2, so 1600/(u^2) = 4 + ((u^4)/400). This equation is equivalent to(U^3) +1600.U -640000 =0 or (U - 80).((U^2) +80.U +8000) = 0 with U = u^2, U = 80 is the unique solution, then u = sqrt(80) = 2.sqrt(20) We then have BP = (u^2)/20 = 80/20 = 4. Now in triangle ABC: AC^2 = BA^2 + BC^2 = (40/u)^2 + (u^2) = 1600/80 + 80 = 100, so AC = 10 The triangle BQC is isosceles, Q is on the mediator of [B, C], so it is also the middle of [A, C], so QC = AC/2 = 10/2 = 5 and PQ = AC - AP - QC = 10 - 2 - 5 = 3. Finally the right triangle BPQ has PQ = 3 for basis and BP = 4 for height, its area is (1/2).3.4 = 6
Triangle ABC's area is given as 20 at the puzzle's start. If the area of ABC is 20, how come it is assumed that the triangle BPC is 20 too if BPC is contained in ABC at 1:45?
For an equation of b^3 + 4b - 80 = 0, we can easily think a value of 4 as a value of the "b". Yes, 4 x 4 x 4 + 4 x 4 - 80 = 0. But, what a very-non-mathematical way!!! If we get an equation of b^3 + 4b - 81 = 0. Then, how can you solve the problem? Answer me!
I think there is a logic error in this solution. 1. The red triangle is also a rectangle triangle. 2. If you draw a perpendicular from Q to the baseline, you will get two more rectangular triangles. 3. All the black triangles are congruent. 4. The red triangle got the same height and the same hypothenuse, so also the second cathede must be the same, with means 2. 5. Your triangle would only be possible, if your hypothenuse would have a kink at P. 6. There are 4 congruent triangles. So all we have to do is dividing 20 through 4. That is 5.
your 3. is false: APB is only similar, not congruent with the other two black triangles. I found that if BQ = QC = x then AB = 2√ x. Tracing QM perpendicular to BC we can see that QM = AB/2 then doing AB : QC = AP : QM AB : x = 2 : AB/2 AB²/2 = 2x AB = 2√ x
Easily prove that triangle ABQ is isosceles. Let AQ=QB=QC=x. The area of triangle ABC is AC*BP/2=20. Therefore, AC*BP=40. Therefore, BP=20/x. Applying the Pythagorean theorem to triangle BPQ, we find x²=(x-2)²+(20/x)². This is equivalent to x³-x²-100=0. That is, (x-5)(x²+4x+20)=0. Therefore, x=5. Since PQ=3 and BP=4, the area of triangle BPQ is (3*4)/2=6.
Small shortcut: Q is the center of the circle passing thru A, B and C and AC is the diameter. So c = a + 2.
I appreciate for your daily videos. Thank you for your time.
@13:36 that c=2+a is easy to see from the equality QC and QB. If we drop a perpendicular on BC we will have OM as a middle line of a triangle which is equal to a half of a base AB.
At about 7:38, Math Booster makes an educated guess that b = 4 is a root. The root can be found without guessing by doing a half interval search. Start the search with 2 values of b, one where b³ + 4b - 80 is negative and one where it is positive. Since b³ + 4b - 80 is continuous, there must be a zero crossing, a value of b where b³ + 4b - 80 is zero, somewhere between our two values. We know that as b increases in absolute value, b³ will dominate, so there must be negative values of b which produce a negative value of b³ + 4b - 80 and positive values of b which produce a positive value of b³ + 4b - 80. We try b = -10 and find b³ + 4b - 80 = -1040 and, for b = 10, b³ + 4b - 80 = 960. So, try -10 and 10 as limits. We start by testing b = 0 (half way between -10 and +10). If positive, we'll go half way to -10 to try our next number. If negative, we go half way to 10. We cut the increment to half. So, after finding that b = 0 produces a negative value, we'll try b = 5. The next try will be either 5 - 2.5 or 5 + 2.5. So, here is a computer program written in python to find b where b³ + 4b - 80 is approximately 0:
# search range is b = -10 to +10
# start at the middle, b = 0
b = 0.0
db = 5.0
# define function to compute delta = ∆
def compute_delta(b):
# ∆ = b³ + 4b - 80, rewrite in python notation
delta = b**3 + 4*b -80
return(delta)
for n in range(0,51): # after 50 divisions by 2, db is sufficiently small
delta = compute_delta(b)
if (delta > 0):
b = b - db
else:
b = b + db
db = db/2
b = round(b , 15) # round b to 15 decimal places
formatted = "{:.15f}".format(b)
print(f"b = ",formatted)
Output:
b = 3.999999999999999
=== Code Execution Successful ===
So, we round up to 4, try b = 4 and find that b³ + 4b - 80 is exactly 0. We can factor out (b - 4) to get a quadratic, which we know how to solve. In this case, the roots of the quadratic are not real numbers, so b = 4 is the only real root.
One viewer came up with a solution that required finding roots for x³ -x² -100. The python program can be rewritten to solve x³ -x² -100 = 0 for x by replacing x with b and changing one line. The line delta = b**3 + 4*b -80 becomes delta = b**3 - b**2 - 100:
Output
b = 5.000000000000004
=== Code Execution Successful ===
Round down to 5 and try b = 5.
An afterthought, refining the limits: we know that b or x must be positive, so we could try a lower limit of 0, which produces a negative value for our third order expression. Choose an upper value of b or x which produces a positive value.
Thank you.
Since we have a right triangle and QC = QB then QC = QA. Let's Call each of these "x".
Area of △ABQ = 10 so hx = 20. Also 2(2x - 2) = h², and since hx = 20, h²x² = 400.
So 2x²(2x - 2) = 400 and x²(x - 1) = 100 from which x = 5 so h = 4.
With this A = h(x - 2)/2 = 6.
Let u = BC, then BA = (2.area of ABC)/u = 40/u. Triangles APB and ABC are similar, so AP/BP = AB/BC = (u^2)/40, so 2/BP = 40/(u^2)
we then have that BP = (u^2)/20. Now in triangle APB: AB^2 = AP^2 + BP^2, so 1600/(u^2) = 4 + ((u^4)/400).
This equation is equivalent to(U^3) +1600.U -640000 =0 or (U - 80).((U^2) +80.U +8000) = 0 with U = u^2, U = 80 is the unique solution,
then u = sqrt(80) = 2.sqrt(20)
We then have BP = (u^2)/20 = 80/20 = 4. Now in triangle ABC: AC^2 = BA^2 + BC^2 = (40/u)^2 + (u^2) = 1600/80 + 80 = 100, so AC = 10
The triangle BQC is isosceles, Q is on the mediator of [B, C], so it is also the middle of [A, C], so QC = AC/2 = 10/2 = 5
and PQ = AC - AP - QC = 10 - 2 - 5 = 3. Finally the right triangle BPQ has PQ = 3 for basis and BP = 4 for height, its area is (1/2).3.4 = 6
Triangle ABC's area is given as 20 at the puzzle's start. If the area of ABC is 20, how come it is assumed that the triangle BPC is 20 too if BPC is contained in ABC at 1:45?
AC*BP/2 is [ABC]
@@rodicabrudea923 Got it! Thank you!
You have not disclosed some information. You have only brought these information during your demonstration.
For an equation of b^3 + 4b - 80 = 0, we can easily think a value of 4 as a value of the "b". Yes, 4 x 4 x 4 + 4 x 4 - 80 = 0. But, what a very-non-mathematical way!!! If we get an equation of b^3 + 4b - 81 = 0. Then, how can you solve the problem? Answer me!
I think there is a logic error in this solution.
1. The red triangle is also a rectangle triangle.
2. If you draw a perpendicular from Q to the baseline, you will get two more rectangular triangles.
3. All the black triangles are congruent.
4. The red triangle got the same height and the same hypothenuse, so also the second cathede must be the same, with means 2.
5. Your triangle would only be possible, if your hypothenuse would have a kink at P.
6. There are 4 congruent triangles. So all we have to do is dividing 20 through 4. That is 5.
How do you know these triangles are congruent?
No. Triangle ABP has area 4, not 5.
your 3. is false: APB is only similar, not congruent with the other two black triangles. I found that if BQ = QC = x then AB = 2√ x. Tracing QM perpendicular to BC we can see that QM = AB/2 then doing
AB : QC = AP : QM
AB : x = 2 : AB/2
AB²/2 = 2x
AB = 2√ x
6 is what I got (after a few false starts).