Chinese Math Olympiad Problem | A Very Nice Geometry Challenge
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- čas přidán 6. 03. 2024
- Chinese Math Olympiad Problem | A Very Nice Geometry Challenge
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You always avoid having to solve a fourth degree equation! Bravo for this solution!
Thanks for providing a solution to this problem.
I saved an almost identical problem [as a photo] a few years ago; but somehow I didn’t save the solution or a link to the solution. I gave up on the problem; thinking something was missing from the ‘illustration.’ So I was happy that you provided an approach and a solution that had been, sadly, just beyond my ken. Thanks again!
Much easier way:
ΔBEF ||| ΔADF (= ∠s)
∴ x / (x+1) = √(1-x²) / 1
Solve for x
A more brute-force attack: It's relatively easy to remove all "a" and end up with a quartic equation in x only:
x⁴ + 2x³ + x² - 2x - 1 = 0.
Now try to make the LHS a perfect square, by adding 2x² + 4x + 2 to both sides, such that the coefs on LHS is the nicer {1,2,3,2,1}:
x⁴ + 2x³ + 3x² + 2x + 1 = 2x² + 4x + 2 = 2 (x+1)².
Since {1,2,3,2,1} is the convolution of {1,1,1} with itself, the LHS is (x² + x + 1)². This leads to
(x² + x + 1)² = 2 (x+1)²
After taking sqrt, we see that all we need to solve is
x² + x + 1 = ±√2 (x+1),
two quadratics that are now tractable, i.e.,
x² + (1±√2) x + 1±√2 =0.
By Decartes' rule of signs, if all coefs are >0, there couldn't be any positive roots. So the equation with "+" sign can be dropped and only one quadratic needs to be solved, i.e.,
x² + (1-√2) x + 1-√2 =0.
I tried 4 hours to solve that equation and I failed. Bravo !
Thanks for the comment. Actually my method is based on some luck as well. When I tried to make LHS look nicer, I hadn't expected the RHS also look that nice. @@mariopopesco
This ought to amuse you.
Let's call it "a numerical programmer's brute force approach" ;-)
#!/usr/bin/perl; # on a MAC;
my $k;
my $dk = 0.1;
for( $k = $dk; $k < 3; $k += $dk )
{
my $f = $k**4 + 2 * $k**3 + $k**2 - 2 * $k - 1;
my $h = $k / (1 + $k);
next if $f < 0;
printf "k %.9f ", $k;
printf "f %.9f ", $f;
printf "
";
$k -= $dk; $dk /= 2.7182818; last if $dk < 1e-11; # My special loop magic.
}
my $g = ((sqrt(2) - 1) + sqrt( 2 * sqrt(2) - 1 )) / 2;
printf "video = %.9f
", $g;
---------- OUTPUT ----------------------------------------------------------------------
k 0.900000000 f 0.124100000
k 0.910363834 f 0.203832074
k 0.887109418 f 0.028300022
k 0.883533303 f 0.002376546
k 0.884049288 f 0.006099571
k 0.883565313 f 0.002607343
k 0.883387269 f 0.001323919
k 0.883230582 f 0.000195023
k 0.883206486 f 0.000021467
k 0.883209963 f 0.000046508
k 0.883206702 f 0.000023021
k 0.883203832 f 0.000002351
k 0.883204005 f 0.000003597
k 0.883203617 f 0.000000800
k 0.883203557 f 0.000000370
k 0.883203535 f 0.000000212
k 0.883203516 f 0.000000072
k 0.883203509 f 0.000000021
k 0.883203506 f 0.000000002
k 0.883203506 f 0.000000003
k 0.883203506 f 0.000000001
k 0.883203506 f 0.000000000
k 0.883203506 f 0.000000000
k 0.883203506 f 0.000000000
video = 0.883203506
... thus showing that the solution found by successive approximation converges relatively quickly to the exact form given in the video (last bit above)
Nicely done.
I used an intermediate method going through trigonometry letting x=cos(thêta), BE=sin(thêta) and half-angle formulas with the classic change of variable t=tan(thêta/2) which still gives a quartic equation but a simpler one. Getting the t value needs after some tricky calculation to get back to the x world.
Anyway this problem is interestlng as we have here in the discussions and the video different ways to get the solution.
The brute force way with the full degree terms quartic equation.
The intermediate way I used, leading to a simpler quartic equation to solve (After several tries I must admit but without looking at the presented solution)
And finally, the nicer way (avoiding any quartic equation to solve) of the youtube channel owner.
Greetings and congrats to all.
P.S: You can see above my detailed solution.
Triangle DCE simular triangle EBF, ^DEC=^BEF.
DF - DE = 1
Knowing that angles CDE = DFA = α, we can see that:
DF = 1/sin α
DE = 1/cos α
1/sin α - 1/cos α = 1
(cos α - sin α)/sinαcosα = 1
cos α - sin α = sinαcosα multiplying by 2
2cos α -2 sin α = sin2α doing the square
4cos²α + 4sin²α - 4*2sinαcosα = sin²2α
sin²2α + 4sin2α - 4 = 0 and solving the quadratic equation:
sin 2α = 2√ 2 - 2 = 2(√ 2 - 1)
2sinαcosα = 2(√ 2 - 1)
setting cos α = x and sinα = y we ca solve the system:
xy = √ 2 - 1
x² + y² = 1
x²y² = 3 - 2√ 2 (doing the square)
y² = 1 - x²
x²(1 - x²) = 3 - 2√ 2
x² - x⁴ = 3 - 2√ 2
setting x² = t we have:
t² - t + 3 - 2√ 2 = 0
x² = t = 1 + √ (8√ 2 - 11)/2
cos α = √ (1 + √ (8√ 2 - 11)/2)
BF = 1 * cos α = √ (1 + √ (8√ 2 - 11)/2) = 0,8832...
Similar triangles + Pythagorean theorem
Thank you for this nice problem.
I found it was more of an algebra problem than a geometry one.
Here is the way I solved it in a detailed way and I hope you will tell me if you like it.
Of course I didn't look at the solution before solving it so it may seem the same way of yours.
Greetings.
Recall of the problem
ABCD is a square of side length equals to 1.
Point E is on segment BC.
DE intersects AB at point F,
such that length EF equals to 1.
Determine the length x=BF.
Let's start:
ABCD being a square of side length 1,
we have
(AD) // (BE) // (BC)
So (ADF) and (BEF) are similar triangles
then BE/BF=DA/AF. And if we let BE=y,
we have y/x=1/(x+1) or
yx+y=x (i)
As (BFE) is a rectangle triangle in B with EF=1,
then 0
If x=sin(a), then the equation is: sin(a)+tg(a)=1
If t=tg(a/2) then sin(a)=2t/(1+t^2), tg(a)=2t/(1-t^2) and the equalition is: t^4+4t-1=0
After some steps: (t^2-sqrt(2)t +1 - sqrt(2)) (t^2+sqrt(2)t +1 + sqrt(2))=0
The first line has a mistake: If x=cos(a), ...
Very nice solution ❤
Yes ❤
So what is "x" ? + keep getting .8944 or thereabouts. What is angle EFB?
0.8832...EFB=arctg(1/1,8832)=27,98
Thank you. Very close to 26.56 degrees, which is what I got. @@giuseppemalaguti435
Although this is a geometry problem, the difficult part is actually the tricky algebra involved.
🎯
Make an integrated formula for🇮🇳 it⌚💻 so it becomes universal by nature🌿🍃😅😅
.5×1.=.5÷2=.25
جسب تطابق المثلثات فأن ..
س= ١ .
2 26 12 2/12 1/2 1 0