Geometry Challenge

01092024, 05:15 PM
(This post was last modified: 01092024 06:03 PM by Albert Chan.)
Post: #21




RE: Geometry Challenge
Trig (numerical) solution
Solve by Casio FX3650P Triangle Solver (can handle both SSS, SAS) Note: this calculator implied multiply has higher precedence than ×/÷ P3 Wrote:?→A: ?→B: ?→C: BC→Y: Program is most accurate if side A is the smallest. see https://www.hpmuseum.org/forum/thread11...#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) 

01092024, 05:39 PM
Post: #22




RE: Geometry Challenge  
01112024, 01:06 PM
Post: #23




RE: Geometry Challenge
(01092024 06:18 AM)Johnh Wrote: i reckon the angle BAD is 85 degrees exactly. You are so close to getting exact solution! (below, angles in degree) Code: (1 + cos10  cos70) = (1 + sin40) 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 

01112024, 09:38 PM
Post: #24




RE: Geometry Challenge
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 oldschool mathemagic! 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) 

01132024, 02:00 PM
Post: #25




RE: Geometry Challenge
(01112024 09:38 PM)Johnh Wrote: Those trig equivalencies are like oldschool mathemagic! 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. (01092024 03:09 PM)Thomas Klemm Wrote: 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 

01132024, 08:15 PM
Post: #26




RE: Geometry Challenge
Anyone else try CoGo? (see post #20)
BEST! SlideRule 

01132024, 09:49 PM
Post: #27




RE: Geometry Challenge  
01132024, 10:15 PM
Post: #28




RE: Geometry Challenge
(01132024 09:49 PM)Thomas Klemm Wrote:(01132024 02:00 PM)Albert Chan Wrote: … = cis(10) * (1  cis(60) = 1 + cis(120) = cis(60)) = … 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) 

01132024, 10:35 PM
Post: #29




RE: Geometry Challenge
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. 

01142024, 01:16 PM
(This post was last modified: 01142024 01:42 PM by Albert Chan.)
Post: #30




RE: Geometry Challenge
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  (180170)/2 = 65 ∠BAO = (18070)/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 

01152024, 01:19 AM
Post: #31




RE: Geometry Challenge
For those with Casio fx9750GIII 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 

01212024, 04:08 AM
Post: #32




RE: Geometry Challenge
(01152024 01:19 AM)paul0207 Wrote: For those with Casio fx9750GIII or equivalent, I ran the problem with the Geometry Application, 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! 

« Next Oldest  Next Newest »

User(s) browsing this thread: