Geometry Challenge

[Albert Chan]

Trig (numerical) solution

Solve by Casio FX-3650P Triangle Solver (can handle both SSS, SAS)
Note: this calculator implied multiply has higher precedence than ×/÷

P3 Wrote:?→A: ?→B: ?→C: B-C→Y:
-A≥0 ⇒ √(4BC(sin A⅃2)²+Y²→A◢
(A+Y)(A-Y)÷4→Y: √(Y(BC-Y→D◢
Y/BC: 2 sin^-1√Ans→X◢
A-C: (B+Ans)(B-Ans)/4AC: 2 sin^-1√Ans→Y◢
π^r-Y-X

Program is most accurate if side A is the smallest.
see https://www.hpmuseum.org/forum/thread-11...#pid107487

ΔBAD (sides AD,AB,BD = Program sides A,B,C), degree mode

[Prog] 3
A? -65 EXE     // negative side means angle between B,C
B? 1 EXE
C? 2 sin(170/2
EXE --> A = 1.812615574
EXE --> D = 0.902859012 (area)
EXE --> X = 64.99999999 (∠A)
EXE --> Y = 30 (∠B)
EXE --> 85.00000001 (∠C)

[Thomas Klemm]

(01-09-2024 03:09 PM)Thomas Klemm Wrote:  Consider \(B\) the origin of \(\mathbb{C}\).

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

Now you have to either tilt your head or your smartphone:

1 ENTER
1 ENTER 110 COMPLEX +
1 ENTER 120 COMPLEX +

1.81261557407 ∠ 85

[Albert Chan]

(01-09-2024 06:18 AM)Johnh Wrote:  i reckon the angle BAD is 85 degrees exactly.
https://www.hpmuseum.org/forum/attachment.php?aid=13154

You are so close to getting exact solution! (below, angles in degree)

Code:

          (1 + cos10 - cos70) = (1 + sin40)
    _____A__________E
    70  / ＼ θ    90|
       /    ＼      |
      /       ＼    | (sin70 - sin10) = cos40 = sin50
     /          ＼  |
    /             ＼|
   / 70             D
B  -------- C

AD^2 = (1+sin40)^2 + cos40^2 = 1 + 2*sin40 + 1 = 4*(1+cos50)/2 = (2*cos25)^2
AD = 2*cos25

sinθ = DE / DA = (2*sin25*cos25) / (2*cos25) = sin25      → θ = 25
∠BAD = 180 - 70 - 25 = 85

[Johnh]

Thanks Albert

Once id seen that nice round 85 degree result, I suspected there'd be an exact formula solution, and I thank you for showing it.

Those trig equivalencies are like old-school mathe-magic! Despite a lifetime of working things out, I haven't seen them directly in 45 years since high school.

Also, since this is a calc forum, I was quite impressed with the triangle solver on my HP39GII, once squinted at in order to read it. It can solve with any three of 3 sides and 3 angles, (so long as it gets at least one side)

[Albert Chan]

(01-11-2024 09:38 PM)Johnh Wrote:  Those trig equivalencies are like old-school mathe-magic!

If I were doing this problem in a exam, I would use vectors, instead of trigonometry.
Solving trig problem may need addition/removal of lines in sketch, which take time.

With vectors, we could do geometric tricks with algebra. (swapping lines is same as swapping terms)

Also, identities may be derived geometrically.
Isosceles triangle (2 sides = 1, exterior angle = 2θ), we have:

1 + cis(2θ) = 2*cos(θ) * cis(θ)

Getting this with trig identities is a bit more messy.

1 + cis(2θ) = (1+cos(2θ), sin(2θ)) = (2*cos(θ)^2, 2*sin(θ)*cos(θ)) = 2*cos(θ) * cis(θ)

BTW, your trig setup is equivalent to Thomas Kleem's vector setup.

(01-09-2024 03:09 PM)Thomas Klemm Wrote:  

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

Then:

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

Solve with vectors (angles in degree)

cis(10) - cis(70) = cis(10) * (1 - cis(60) = 1 + cis(-120) = cis(-60)) = cis(10 - 60) = cis(-50)

vec(AD) = 1 + cis(-50) = 2*cos(-25) * cis(-25)

|AD| = 2*cos(25) ≈ 1.8126
∠BAD = 180 - 70 - 25 = 85

[SlideRule]

Anyone else try CoGo? (see post #20)

BEST!
SlideRule

[Thomas Klemm]

(01-13-2024 02:00 PM)Albert Chan Wrote:  … = cis(10) * (1 - cis(60) = 1 + cis(-120) = cis(-60)) = …

You lost me somewhere between here and the missing opening parenthesis.

---

BTW: I'm still busy drawing owls.

[Albert Chan]

(01-13-2024 09:49 PM)Thomas Klemm Wrote:  

(01-13-2024 02:00 PM)Albert Chan Wrote:  … = cis(10) * (1 - cis(60) = 1 + cis(-120) = cis(-60)) = …

You lost me somewhere between here and the missing opening parenthesis.

Sorry about the shorthand (to make above clearer, I added matching parenthesis in red)

cis(10) - cis(70)
= cis(10) * (1 - cis(60))
= cis(10) * (1 + cis(-120)) 
= cis(10) * cis(-60)
= cis(10 - 60)
= cis(-50)

[Thomas Klemm]

Oh, well.
Talking about bending notation, it reminds me of students chaining calculations:

\(
5 + 8 = 13 - 3 = 10 \div 5 = 2 \cdots
\)

That's not how equations work.

[Albert Chan]

There are many ways to solve the puzzle, here is another. (again, angles in degree)
Let Intersection Point of a Quadrilateral of ABCD be O

∠ABO = 70 - (180-170)/2 = 65
∠BAO = (180-70)/2 = 55
∠AOB =180 - 65 - 55 = 60
∠COD = ∠AOB = 60            // opposite angles
∠OCD = 170 - 55 = 115

Law of sines:
ΔAOB: OA/sin65 = 1/sin60
ΔCOD: OD/sin115 = 1/sin60

Since sin65 = sin115, ΔAOD is isosceles

∠OAD = ∠ODA = ∠AOB / 2 = 30
∠BAD = ∠BAO + ∠OAD = 55 + 30 = 85

OA = OD = sin65/sin60 = cos25/cos30
AD = 2 * OA * cos(∠OAD) = 2 * cos25

[paul0207]

For those with Casio fx-9750GIII or equivalent, I ran the problem with the Geometry Application,
   
Please find Geometry file attached and zipped (I couldn't upload directly as g1m).

Paul

[richmit]

(01-15-2024 01:19 AM)paul0207 Wrote:  For those with Casio fx-9750GIII or equivalent, I ran the problem with the Geometry Application,

Paul

I totally feel this! The geometry application on the Casio Classpad II has become my prime tool for these kinds of geometry problems. It's ridiculously easy to simply draw the diagram with the stylus on the touch screen, enter the known quantities, and read off the automagically found unknowns. It almost feels like cheating!