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Comparing angles θ and φ

I’m considering whether θ can always be less than φ. Let’s assume r equals b and c. After some algebra, it turns out that for φ = 45°, cos θ ends up being greater than cos φ, meaning θ will always be smaller than φ in this symmetric case.

Testing extreme cases for unequal r

I’m now testing if θ can be greater than φ when b and c are unequal. To check this, I use a formula for cos θ and compare it to cos φ. After expanding the inequality, I’ll see if there's any chance cos θ could be less than cos φ, which would imply that θ > φ.

Evaluating the inequality for θ > φ

I’m expanding the inequality. After simplifying, the condition becomes: D > 0, which would mean that cos θ < cos φ. I then rewrite it, getting an expression involving p, q, and h^2. Next, I'll use h^2 as a substitution to proceed and check if the condition holds.

Determining when condition holds

The condition becomes h^2[(p^2 + q^2) - 2√2 pq] > h^4. I simplify it to an expression involving p and q. It turns out that when p ≈ q, the result is negative. However, if the ratio p/q is large enough, the expression can become positive. We find that E is positive when t > 2.414 or t < 0.414.

Finalizing counterexample and solution

If the ratio of p to q exceeds 2.414 or is below 0.414, then angle θ can exceed 45°. I picked p = 3, q = 1, and h = 1 to show that the angle can be larger than 45°. With this concrete counterexample, it’s clear that the angle doesn’t always have to be less than 45°.

Thought for 1m 23s

Exploring geometry logic without counterexamples

The user wants to know if we can prove whether ∠BAC is less than 45° using only plane geometry, without counterexamples. I could suggest a synthetic approach. We could show how the angle depends on the distances OB, OC, AO, and angle BOC, using projections and right triangles to form a conceptual argument.

Exploring angle variation within planimetry

I’m thinking we can show that ∠BAC isn’t fixed by using a sphere centered at A, with B and C on the plane. By fixing AO and angle BOC, the angle BAC can vary, potentially approaching 0 or 180°. Using the law of cosines in triangles, we can prove this by working with algebraic expressions within plane geometry.

Reasoning about angle behavior with limits

I’m considering showing that the function f(p,q) can be less than cos45 for certain values. It would still be an existence proof, without needing explicit numbers. We can reason by taking limits—like when q approaches 0 or p becomes very large, adjusting side lengths and angles to approach a desired configuration.

Approximating angle behavior in triangle ABC

I’m considering a triangle ABC where AB is nearly equal to BC. The angles at A become close to the angle between AB and AC when BC ≈ AB. Using the law of cosines, we can approximate that as AB becomes much larger than AC, angle A tends toward 90°, which could exceed 45°.

Building a geometric argument with limits

I’m working on showing that angle BAC can be made larger than 45° by constructing inequalities. Using a derived formula, if p becomes very large and q small, we can show that cosA becomes smaller than cos45, which leads to an angle greater than 45°. Since the argument is geometric, I’m focusing on using the cosine theorem and limit analysis.

Explaining angle variation in triangle geometry

I’ve derived the formula for cos(∠BAC) using the law of cosines. By fixing certain values and varying others, I can show that the angle ∠BAC can exceed 45°, depending on how far point B is from O. The limit argument gives us a rigorous way to prove this result by comparing coefficients as B moves. This shows how angle BAC can be adjusted, even reaching angles close to 90°.

Thought for 1m 0s

Considering geometric reasoning for angle calculation

I need to focus purely on geometric reasoning. Looking at the cross-section through point A, I realize that AB forms the hypotenuse of a right triangle with AO and OB as legs. Similarly, for the other plane, the angle between planes depends on the angle between lines OB and OC. This helps me relate the angle between AB and AC to various lengths.

Exploring intuition for triangle geometry

The idea is that if B is far from O along a 45-degree angle with OC, then AB will be roughly equal to OB, and BC will approach OB (because in triangle BOC, as OB increases, BC remains about the same length). With AC fixed, triangle ABC becomes nearly isosceles, with small angle at B.

Clarifying angle relationships in triangle ABC

If AB ≈ BC and AC is small, the angle at B will be small, and angles at A and C approach 90. To make sure angle A is greater than 45, I need to choose B where AB=BC. That way, angles A and C are equal, and if AB is large, angle B becomes smaller, keeping angle A > 45.

Finding the position of B geometrically

To prove such a B exists, I'll approach this geometrically. I’ll define ray l from O, making a 45° angle with OC on the base plane. As point B moves along this ray, I’ll compare AB and BC as functions of OB and use monotonicity to show that AB=BC happens at some position, and beyond that AB < BC.

Exploring angle relationships with geometric reasoning

I’m comparing AB and BC, starting with AB² - BC² = h² - c² + 2tc cos(45°). As t increases, this expression eventually becomes positive. If c > h, the result at t=0 is negative, but as t grows, it crosses zero, ensuring AB=BC at some point. By adjusting C and B along specific rays, angle A will be greater than 45°.

Adjusting base and apex angles based on triangle constraints

To ensure an apex angle <90, we need to choose B far enough so that AB becomes large. As B moves, we want AB = BC concurrently. If AB = BC happens too late, we adjust the selection of C to make this work. I'll explore if ensuring c < h can help in this situation.

Ensuring angle A >45 by adjusting base and side length

If h > c, then AB > BC at the start and AB can't equal BC as t increases. For AB = BC, we need c > h, but this causes base AC to become larger than both sides. To make angle A > 45, we need to ensure that AC < √2 AB.

Choosing configuration to balance sides properly

We want a configuration where AB is slightly larger than AC. By choosing c just a bit larger than h, we keep t0 small, which helps balance AB and AC. In some examples, AB and BC are nearly equal but large, while AC stays moderate, ensuring angle A is close to 90.

Ensuring triangle inequalities hold for large sides

We have AB and BC being large and roughly equal, while AC remains fixed and small. This makes the triangle nearly isosceles. Now, I need to turn this into rigor using inequalities: if AB and BC are greater than AC, I can use the cosine law to show angle B is acute. With large AB, the angles at A and C are just over 45 degrees.

Ensuring triangle ABC has the desired properties

AB and BC approach equality as we make b larger, and they both grow at the same rate. To ensure angle B is less than 90, we need AB and BC to each be larger than AC. Using the law of cosines, this ensures cos B remains positive. Afterward, we can ensure angle A exceeds 45 degrees.

Thought for 2m 7s