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

Fri Feb 20 15:29:29 PST 2026

 So, you don't remember in High School geometry class, being drilled about the 30-60-Right triangle? I had the answer immediately based on that knowledge. Now, if they had asked the length of the hypoteneuse of that triangle, I would have had to look up a formula. 
-- Bob Primak 

    On Friday, February 20, 2026 at 05:08:10 PM EST, Peter Albin via LCTG <lctg at lists.toku.us> 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

 ===============================================
::The Lexington Computer and Technology Group Mailing List::
Reply goes to sender only; Reply All to send to list.
Send to the list: LCTG at lists.toku.us      Message archives: http://lists.toku.us/pipermail/lctg-toku.us/
To subscribe: email lctg-subscribe at toku.us  To unsubscribe: email lctg-unsubscribe at toku.us
Future and Past meeting information: http://LCTG.toku.us
List information: http://lists.toku.us/listinfo.cgi/lctg-toku.us
This message was sent to bobprimak at yahoo.com.
Set your list options: http://lists.toku.us/options.cgi/lctg-toku.us/bobprimak@yahoo.com

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.toku.us/pipermail/lctg-toku.us/attachments/20260220/136acf75/attachment.htm>
-------------- next part --------------
A non-text attachment was scrubbed...
Name: tVV7gluCWSwLiDgr.png
Type: image/png
Size: 105073 bytes
Desc: not available
URL: <http://lists.toku.us/pipermail/lctg-toku.us/attachments/20260220/136acf75/attachment.png>