[Lex Computer & Tech Group/LCTG] AI as a visual geometry problem solver
S Ou
shukong at gmail.com
Sat Feb 21 17:12:40 PST 2026
Maybe throw the problem at this
https://phys.org/news/2026-02-ai-tonggeometry-generates-olympiad-geometry.html
?
On Fri, Feb 20, 2026 at 6:38 PM Robert Primak via LCTG <lctg at lists.toku.us>
wrote:
> Jon,
>
> The angles are right angles. The base line however, is deliberately not
> drawn to scale. The 50-degree angle on the left insures that the base line
> cannot be cleanly bisected, if there is in fact a single point at the apex
> of the diagram.
>
> I generally do geometry spatially, that is by seeing physical objects in
> space. I don't do well using formulas and algorithms. I'm a
> tactile-kinesthetic learner (hands-on).
>
> -- Bob Primak
>
>
> On Friday, February 20, 2026 at 06:28:45 PM EST, Jon Dreyer via LCTG <
> lctg at lists.toku.us> wrote:
>
>
> Oh my goodness! That's a long way to go for mostly wrong answers.
> Congratulations to ChatGPT (and Peter)!
>
> I didn't have the patience to wade through the LaTeX to see if and how the
> others justified or ignored different base angles in an isosceles triangle.
>
> I'm guessing that they assumed that the seemingly vertical line was
> perpendicular to the seemingly horizontal line, and then kept on churning
> until there was enough AI slop to seemingly justify the cognitive
> dissonance.
>
> --
>
> Jon "It's All About That Base" Dreyer
> Math Tutor/Computer Science Tutor <http://www.passionatelycurious.com>
> Jon Dreyer Music <http://music.jondreyer.com>
>
>
> On 2/20/26 16:02, Peter Albin via LCTG wrote:
>
> I had this image show up on my phone and I thought to give ChatGPT ,
> Gemini and Claude a chance to flex their muscles ...
>
> The correct answer is below the responses
>
> They all eventually achieved the correct answer with a fair amount of
> course correcting. These are their initial responses:
>
> *Claude*
> Solving for X
>
> *Key observations from the diagram:*
>
> - A triangle with angles 50° (bottom left) and 60° (bottom right)
> - A vertical line from the apex to the base, with tick marks showing
> the base is bisected (midpoint)
> - X is the angle between the right side of the triangle and the
> vertical line
>
> *Step 1: Find the apex angle of the triangle*
>
> Sum of angles in a triangle = 180° Apex angle = 180° - 50° - 60° = *70°*
>
> *Step 2: Use the exterior angle / angle bisector property*
>
> The vertical line bisects the base (shown by tick marks), but it's *not*
> the angle bisector. I need to find angle X using the triangle formed on the
> right side.
>
> In the right sub-triangle (with the vertical line):
>
> - Bottom right angle = 60°
> - The vertical line meets the base at 90° → bottom middle angle = 90°
> - So the top angle of the right sub-triangle = 180° - 60° - 90° = *30°*
>
> *X is the angle between the median and the right side of the triangle*,
> which equals:
>
> Apex angle - right sub-triangle top angle = 70° - 30° = *but wait...*
>
> The right sub-triangle's top angle IS x, so:
>
> *X = 180° - 90° - 60° = 30°*
> Answer: *B) 30°*
>
> This geometry puzzle is a classic "hard" problem because it requires more
> than just basic angle-sum rules—you have to use the Law of Sines to find
> the solution.
>
> Let’s break it down step-by-step.
> 1. Label the Triangle
>
> Let's assign labels to the vertices to make the math easier:
>
> -
>
> *$A$*: The bottom-left vertex ($50°$).
> -
>
> *$B$*: The bottom-right vertex ($60°$).
> -
>
> *$C$*: The top vertex.
> -
>
> *$D$*: The point on the base where the vertical line hits.
>
> The image shows two segments marked with double tick marks ($||$),
> meaning they are of equal length. Let’s call this length $k$:
>
> -
>
> $AD = k$
> -
>
> $CD = k$
>
> ------------------------------
> 2. Find the Missing Angles
>
> In $\triangle ABC$, the total angle at the top ($C$) is $180° - 50° - 60°
> = 70°$.
>
> We are looking for angle $x$ (which is $\angle BCD$). This means the
> other part of the top angle ($\angle ACD$) is $70° - x$.
>
> Now, let's look at the two smaller triangles:
>
> -
>
> *In $\triangle ADC$*: The angles are $50°$, $(70° - x)$, and $\angle
> ADC$.
> -
>
> *In $\triangle BDC$*: The angles are $60°$, $x$, and $\angle BDC$.
>
> Note that $\angle ADC$ and $\angle BDC$ form a straight line, so:
>
> $\angle ADC = 180° - \angle BDC$.
> ------------------------------
> 3. Apply the Law of Sines
>
> We will use the Law of Sines: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$.
>
> *In $\triangle ADC$:*
> $$\frac{AD}{\sin(70^\circ - x)} = \frac{CD}{\sin(50^\circ)}$$
>
> Since $AD = CD = k$, they cancel out:
> $$\sin(70^\circ - x) = \sin(50^\circ) \quad \dots \text{(Equation 1)}$$
>
> *Wait!* This implies $70 - x = 50$, so $x = 20$. However, we must check
> if this holds true for the other triangle.
>
> *In $\triangle BDC$:*
>
> The shared side between the two triangles is $CD$. Let's relate $CD$ and
> $BD$ using the Law of Sines in $\triangle BDC$:
> $$\frac{CD}{\sin(60^\circ)} = \frac{BD}{\sin(x)}$$
> $$BD = \frac{k \cdot \sin(x)}{\sin(60^\circ)}$$
> ------------------------------
> 4. Solving for x
>
> Because the line $CD$ is not necessarily a perpendicular bisector (it's
> just a segment of length $k$), we use the trigonometric form of Ceva's
> Theorem or simply equate the shared properties.
>
> When you solve the transcendental equation resulting from the ratio of the
> sides:
> $$\frac{\sin(70-x)}{\sin(50)} = \frac{\sin(x)}{\sin(60-x)}$$
>
> (using the fact that $\angle ADC + \angle BDC = 180°$)
>
> By testing the options:
>
> -
>
> If *$x = 30°$*:
> -
>
> In $\triangle BDC$, if $x=30°$ and $\angle B = 60°$, then the third
> angle $\angle BDC$ must be $90°$.
> -
>
> If $\angle BDC = 90°$, then $\triangle ADC$ must also have a $90°$
> angle at $D$.
> -
>
> In $\triangle ADC$, if $\angle D = 90°$ and $\angle A = 50°$, then
> the top angle must be $40°$.
> -
>
> Does $x=30$ fit? Total top angle was $70°$. $40° + 30° = 70°$. *This
> matches perfectly.*
>
> The Answer
>
> Since $x = 30°$ creates a consistent set of triangles where $CD$ is a
> height that satisfies the $70°$ total vertex angle:
>
> *The correct answer is B) 30°.*
>
> Would you like me to show the full trigonometric proof for why $x$ must
> be 30 without using the multiple-choice options?
>
>
> *ChatGPT *(taking much longer than Claude)
>
> Let A and B be the endpoints of the base, and C the top vertex. Point D is
> on AB such that AD = CD. The base angles are ∠A = 50° and ∠B = 60°. We need
> x = ∠DCB.
>
> 1.
>
> Since AD = CD, triangle ACD is isosceles with equal sides AD and CD.
> Therefore, the base angles at A and C are equal:
> ∠CAD = ∠ACD.
> 2.
>
> ∠CAD is the same as the given angle at A, which is 50°. So:
> ∠ACD = 50°.
> 3.
>
> In the big triangle ABC, the angles sum to 180°:
> ∠ACB = 180° − (50° + 60°) = 70°.
> 4.
>
> The angle at C is split by CD into two parts:
> ∠ACB = ∠ACD + ∠DCB = 50° + x.
>
> So:
> 50° + x = 70°
> x = 20°.
>
> *Note:** I reran the question online with ChatGPT5 and it returned the
> same solution almost instantly. The original query was presented to
> ChatGPT4.*
>
>
> *Gemini* (very fast)
>
> This geometry puzzle is a classic "hard" problem because it requires more
> than just basic angle-sum rules—you have to use the Law of Sines to find
> the solution.
>
> Let’s break it down step-by-step.
> 1. Label the Triangle
>
> Let's assign labels to the vertices to make the math easier:
>
> -
>
> *$A$*: The bottom-left vertex ($50°$).
> -
>
> *$B$*: The bottom-right vertex ($60°$).
> -
>
> *$C$*: The top vertex.
> -
>
> *$D$*: The point on the base where the vertical line hits.
>
> The image shows two segments marked with double tick marks ($||$),
> meaning they are of equal length. Let’s call this length $k$:
>
> -
>
> $AD = k$
> -
>
> $CD = k$
>
> ------------------------------
> 2. Find the Missing Angles
>
> In $\triangle ABC$, the total angle at the top ($C$) is $180° - 50° - 60°
> = 70°$.
>
> We are looking for angle $x$ (which is $\angle BCD$). This means the
> other part of the top angle ($\angle ACD$) is $70° - x$.
>
> Now, let's look at the two smaller triangles:
>
> -
>
> *In $\triangle ADC$*: The angles are $50°$, $(70° - x)$, and $\angle
> ADC$.
> -
>
> *In $\triangle BDC$*: The angles are $60°$, $x$, and $\angle BDC$.
>
> Note that $\angle ADC$ and $\angle BDC$ form a straight line, so:
>
> $\angle ADC = 180° - \angle BDC$.
> ------------------------------
> 3. Apply the Law of Sines
>
> We will use the Law of Sines: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$.
>
> *In $\triangle ADC$:*
> $$\frac{AD}{\sin(70^\circ - x)} = \frac{CD}{\sin(50^\circ)}$$
>
> Since $AD = CD = k$, they cancel out:
> $$\sin(70^\circ - x) = \sin(50^\circ) \quad \dots \text{(Equation 1)}$$
>
> *Wait!* This implies $70 - x = 50$, so $x = 20$. However, we must check
> if this holds true for the other triangle.
>
> *In $\triangle BDC$:*
>
> The shared side between the two triangles is $CD$. Let's relate $CD$ and
> $BD$ using the Law of Sines in $\triangle BDC$:
> $$\frac{CD}{\sin(60^\circ)} = \frac{BD}{\sin(x)}$$
> $$BD = \frac{k \cdot \sin(x)}{\sin(60^\circ)}$$
> ------------------------------
> 4. Solving for x
>
> Because the line $CD$ is not necessarily a perpendicular bisector (it's
> just a segment of length $k$), we use the trigonometric form of Ceva's
> Theorem or simply equate the shared properties.
>
> When you solve the transcendental equation resulting from the ratio of the
> sides:
> $$\frac{\sin(70-x)}{\sin(50)} = \frac{\sin(x)}{\sin(60-x)}$$
>
> (using the fact that $\angle ADC + \angle BDC = 180°$)
>
> By testing the options:
>
> -
>
> If *$x = 30°$*:
> -
>
> In $\triangle BDC$, if $x=30°$ and $\angle B = 60°$, then the third
> angle $\angle BDC$ must be $90°$.
> -
>
> If $\angle BDC = 90°$, then $\triangle ADC$ must also have a $90°$
> angle at $D$.
> -
>
> In $\triangle ADC$, if $\angle D = 90°$ and $\angle A = 50°$, then
> the top angle must be $40°$.
> -
>
> Does $x=30$ fit? Total top angle was $70°$. $40° + 30° = 70°$. *This
> matches perfectly.*
>
> The Answer
>
> Since $x = 30°$ creates a consistent set of triangles where $CD$ is a
> height that satisfies the $70°$ total vertex angle:
>
> *The correct answer is B) 30°.*
>
>
> The correct answer (if you got this far and have not solved it yourself)
> was from ChatGPT and is choice B) 20 degrees.
>
> Peter
>
> ===============================================
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