[Lex Computer & Tech Group/LCTG] AI as a visual geometry problem solver

Fri Feb 20 15:35:51 PST 2026

 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
 Jon Dreyer Music 

 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.

   -     
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.

   -     
∠CAD is the same as the given angle at A, which is 50°. So:   
 ∠ACD = 50°.

   -     
In the big triangle ABC, the angles sum to 180°:   
 ∠ACB = 180° − (50° + 60°) = 70°.

   -     
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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