Geometry Challenge

[Albert Chan]

Problem from Chinese Drama, Heart of Genius, Epsiode 7 (~ 30 minutes in)
https://www.youtube.com/watch?v=6dB8oDVqVGY&t=1800s

Code:

   A
   /
B /
  \
   \____
   C   D

AB = BC = CD = 1, ∠B = 70°, ∠C = 170°, Find ∠A (= ∠BAD)

Bonus Challenge, find |AD| too!

[Werner]

Well, the reply seems to be a nice round number, but for now I can't prove that it is, I merely brute-force calculate it..
Werner

[Maximilian Hohmann]

Hello!

(01-08-2024 04:10 PM)Werner Wrote:  ... but for now I can't prove that it is, I merely brute-force calculate it..

In the video (I had to set my VPN to an IP address in the U.S. to watch it) this is supposedly solved by schoolchildren with paper and pencil only. The narration is chinese only, so I don't know what was said about this task and it's solution.

Regards
Max

NB: If I would have to solve it, I would probably use brute force as well and do it numerically with vector addition.

[C.Ret]

Hi every one...

The sketch draw with ASCII characters by Albert Chan is a bit confusing for me.

Is the following figure a more accurate représentation ?

(01-08-2024 04:30 PM)Maximilian Hohmann Wrote:  NB: If I would have to solve it, I would probably use brute force as well and do it numerically with vector addition.

I would prefer using the Triangle Solver Application of a simple HP Prime. Getting the answers of the challenge and the extra only takes two taps on the (Solve) touchscreen soft-menu's button.

[SlideRule]

Try CoGo - solved it in under two minutes.

BEST!
SlideRule

[Albert Chan]

(01-08-2024 06:55 PM)C.Ret Wrote:  Is the following figure a more accurate représentation ?

Yes! Thanks!

Can you make 1 more, with C flipped along line BD?
(B,A,D) location unchanged --> ∠BAD, |AD| unchanged

Hint: flipped C, we have ∠B = 70° - 10° = 60°

[Johnh]

taking BC as a base line, Id work out the vector AD with its angle. Angle D is 10 degrees more than that. All internal angles of a quadrilateral add up to 360 degrees.

angle A = 360 - B - C - D
?

[C.Ret]

(01-08-2024 08:11 PM)Albert Chan Wrote:  Can you make 1 more, with C flipped along line BD?

Certainly, but I wonder what this may improuve ? Now triangle ABC' is equilateral, but I manage to get a solution with the original isocèle triangle ABC ?


EDIT: Ah! Yes, now I see, AC'D is also an isocèle triangle. That's help a lot.

[Johnh]

When you work it out, it's very interesting to find out what a nice round number angle A appears to be, despite all the sine and cosine functions that can be involved!

[rawi]

Solving with triangle solver of the Prime:
1st: Triangle BAC with BC = b (in Triangle Solver) = 1
BA = a = 1
Angle CBA = C = 70 degrees --> CA = c = 1.14715287
Angle BCA = A = 55 degrees
Angle BAC = B = 55 degrees
2nd: Triangle ACD with AC = a = 1.14715287
CD = b = 1
Angle ACD = C = 170 - 55 = 115 degrees
--> Side AD = c = 1.81261557
Angle BAD = 30 Degrees
Angle BAD + Angle BAC is the angle we searched: 85 degrees

Edit: Albert Chan pointed out in a private message that I had a typo concerning the length of CA. This had led to a wrong result. This is corrected now.

[Johnh]

i reckon the angle BAD is 85 degrees exactly.

see attached below

I also tried it with the triangle solver on my HP39gII, solving BCD first and then BAD

[Thomas Klemm]

With the HP-42S in POLAR mode and the help of C.Ret's picture:

1 ENTER
1 ENTER 10 COMPLEX +
1 ENTER 70 COMPLEX -
LASTX CHS /

1.81261557407 ∠ 85

[Werner]

Well that second picture indeed made it easy to prove that the angle is 85° (60+(180-130)/2).
Werner

[Albert Chan]

(01-09-2024 07:05 AM)Thomas Klemm Wrote:  With the HP-42S in POLAR mode and the help of C.Ret's picture:

1 ENTER
1 ENTER 10 COMPLEX +
1 ENTER 70 COMPLEX -
LASTX CHS /

1.81261557407 ∠ 85

How does this work?
Why the subtraction and division?

With vectors, this is how I solved this.

cis(x) + cis(y)
= (cos(x)+cos(y), sin(x)+sin(y))
= (2*cos((x+y)/2)*cos((x-y)/2), 2*sin((x+y)/2)*cos((x-y)/2)

--> cis(x) + cis(y) = 2*cos((x-y)/2) * cis((x+y)/2)

vec(DA)
= vec(DC) + vec(CB) + vec(BA)
= cis(180°) + cis(170°) + cis(60°)
= cis(170°) + cis(180°) + cis(60°)
= cis(170°) + cis(120°)
= 2*cos(25°) * cis(145°)

|AD| = 2*cos(25°) ≈ 1.8126
∠A = 360° - ∠B - ∠C - ∠D = 360° - 70° - 170° - (180-145)° = 85°

[SlideRule]

Graphic
   
BEST!
SlideRule

[paul0207]

SlideRule,
May I ask which COGO software did you use?
I tried this problem with Q-Cogo (https://q-cogo.com/)

Cheers,

Paul

(01-09-2024 12:54 PM)SlideRule Wrote:  Graphic

BEST!
SlideRule

[Thomas Klemm]

(01-09-2024 12:27 PM)Albert Chan Wrote:  How does this work?
Why the subtraction and division?

Consider \(B\) the origin of \(\mathbb{C}\).

Then:

\(
\begin{align}
A &= (1 \measuredangle 70^{\circ}) \\
\\
C &= 1 \\
\\
D
&= C + (1 \measuredangle 10^{\circ}) \\
&= 1 + (1 \measuredangle 10^{\circ}) \\
\\
\overline{AD}
&= D - A \\
&= 1 + (1 \measuredangle 10^{\circ}) - (1 \measuredangle 70^{\circ}) \\
\\
\overline{AB}
&= B - A \\
&= -A \\
\end{align}
\)

To get the angle between \(\overline{AD}\) and \(\overline{AB}\) we can divide them:

\(
\begin{align}
\frac{\overline{AD}}{\overline{AB}} = \frac{1 + (1 \measuredangle 10^{\circ}) - (1 \measuredangle 70^{\circ})}{- (1 \measuredangle 70^{\circ})}
\end{align}
\)

Since \(\left \| \overline{AB} \right \| = 1\) we also get \(\left \| \overline{AD} \right \|\).

[Albert Chan]

Another vector solution, shape rotated, so that arg(vec(AD)) = ∠BAD

Code:

  D
   \
60° \  120°
----- C -----
 110°  \   
    70° \ 110°
   A-----B ------

vec(AD)
= vec(AB) + vec(BC) + vec(CD)
= cis(0°) + cis(110°) + cis(120°)
= cis(110°) + cis(60°)
= 2*cos((110°-60°)/2) * cis((110°+60°)/2)
= 2*cos(25°) * cis(85°)

[Albert Chan]

(01-08-2024 04:30 PM)Maximilian Hohmann Wrote:  In the video (I had to set my VPN to an IP address in the U.S. to watch it) this is supposedly solved by schoolchildren with paper and pencil only. The narration is chinese only, so I don't know what was said about this task and it's solution.

Kids are competing to get into math olympiad competition.

Exam only care about correct solution, not the process of getting it.
There is no solution shown, which make it more interesting!

I've had similar experience once, get into US chemistry olympiad camp, in Colorado Springs.
What a trip! Everything is Free! Lobsters everyday!

Unfortunately, my hands eyes coordination is bad, and get screened out.

[SlideRule]

(01-09-2024 01:44 PM)paul0207 Wrote:  SlideRule,
May I ask which COGO software did you use?
I tried this problem with Q-Cogo (https://q-cogo.com/)

Cheers,

Paul

The CoGo routines/programs are my own BUT developed from standard algorithms / equations / formulas etc..

BEST!
SlideRule