[Lex Computer & Tech Group/LCTG] AI as a visual geometry problem solver
Jon Dreyer
jon at jondreyer.org
Fri Feb 20 15:50:04 PST 2026
How can the angles (I assume you're talking about the angles made by the
vertical-ish line and the horizontal-ish line) be right angles, when the
angle adjacent to x must be 50° (because it's the other base angle of
the isosceles triangle)?
The reason we need math is that our instincts aren't always reliable.
--
Jon Dreyer
Math Tutor/Computer Science Tutor <http://www.passionatelycurious.com>
Jon Dreyer Music <http://music.jondreyer.com>
On 2/20/26 18:35, Robert Primak via LCTG 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 *$x = 30°$*:
>
> o
>
> In $\triangle BDC$, if $x=30°$ and $\angle B = 60°$, then the
> third angle $\angle BDC$ must be $90°$.
>
> o
>
> If $\angle BDC = 90°$, then $\triangle ADC$ must also have a
> $90°$ angle at $D$.
>
> o
>
> In $\triangle ADC$, if $\angle D = 90°$ and $\angle A = 50°$,
> then the top angle must be $40°$.
>
> o
>
> 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 *$x = 30°$*:
>
> o
>
> In $\triangle BDC$, if $x=30°$ and $\angle B = 60°$, then the
> third angle $\angle BDC$ must be $90°$.
>
> o
>
> If $\angle BDC = 90°$, then $\triangle ADC$ must also have a
> $90°$ angle at $D$.
>
> o
>
> In $\triangle ADC$, if $\angle D = 90°$ and $\angle A = 50°$,
> then the top angle must be $40°$.
>
> o
>
> 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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