CF 234F - Fence
This is a Type B - “Prove that” problem. The statement to prove is: If $$a+b=tanfrac{gamma}{2}(atanalpha+btanbeta),$$ then the triangle is isosceles. The proposed proof attempts exactly this implication. It does not merely prove a weaker statement.
Rating: 1800
Tags: dp
Solve time: 1m 5s
Verified: no
Solution
Problem-Type Check
This is a Type B - “Prove that” problem.
The statement to prove is:
If
$$a+b=\tan\frac{\gamma}{2}(a\tan\alpha+b\tan\beta),$$
then the triangle is isosceles.
The proposed proof attempts exactly this implication. It does not merely prove a weaker statement. The intended conclusion is $a=b$, and the proof derives $\alpha=\beta$ and then uses the sine law to conclude $a=b$.
The proof does cover arbitrary nondegenerate triangles satisfying the hypothesis. No missing cases appear at the level of the overall structure.
However, correctness depends on whether the intermediate trigonometric reductions are valid.
Step-by-Step Verification
Step 1: Substitute $a=2R\sin\alpha$, $b=2R\sin\beta$ using the sine law - VALID
The proof correctly uses
$$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=2R.$$
Substituting into the hypothesis gives
$$\sin\alpha+\sin\beta = \tan\frac{\gamma}{2} \left( \sin\alpha\tan\alpha+\sin\beta\tan\beta \right).$$
This is correct.
Step 2: Replace $\tan(\gamma/2)$ by $\cot\frac{\alpha+\beta}{2}$ - VALID
Since
$$\gamma=\pi-(\alpha+\beta),$$
we have
$$\tan\frac{\gamma}{2} = \tan\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right) = \cot\frac{\alpha+\beta}{2}.$$
Correct.
Step 3: Rewrite
$$\sin\alpha\tan\alpha = \sec\alpha-\cos\alpha$$
- VALID
Indeed,
$$\sin\alpha\tan\alpha = \frac{\sin^2\alpha}{\cos\alpha} = \frac{1-\cos^2\alpha}{\cos\alpha} = \sec\alpha-\cos\alpha.$$
Correct.
Step 4: Derive
$$1-\cos\alpha\cos\beta = \sin^2\frac{\alpha+\beta}{2} + \sin^2\frac{\alpha-\beta}{2}$$
- VALID
Using
$$\cos\alpha\cos\beta = \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}2,$$
we get
$$1-\cos\alpha\cos\beta = 1-\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}2.$$
Also,
$$\sin^2\frac{\alpha+\beta}{2} = \frac{1-\cos(\alpha+\beta)}2,$$
and similarly for the second term. Summing gives exactly the claimed identity.
Correct.
Step 5: Simplify the original equation to
$$\cos\frac{\alpha-\beta}{2} = \frac{\cos\frac{\alpha-\beta}{2}} {\cos\alpha\cos\beta} \cos^2\frac{\alpha+\beta}{2}$$
- UNJUSTIFIED
This is the key reduction of the proof, and the derivation is not adequately shown.
The proof presents several partially overlapping manipulations, abandons one route midway, and then asserts “after cancellation” the desired equation. The intermediate algebra is incomplete.
To verify whether the claimed simplification is actually correct, we compute carefully.
From earlier steps:
$$\sin\alpha+\sin\beta = 2\sin S\cos D,$$
where
$$S=\frac{\alpha+\beta}{2},\qquad D=\frac{\alpha-\beta}{2}.$$
Also,
$$\sin\alpha\tan\alpha+\sin\beta\tan\beta = \frac{(\cos\alpha+\cos\beta)(1-\cos\alpha\cos\beta)} {\cos\alpha\cos\beta}.$$
Using
$$\cos\alpha+\cos\beta = 2\cos S\cos D,$$
the right-hand side of the original equation becomes
$$\cot S \cdot \frac{2\cos S\cos D(1-\cos\alpha\cos\beta)} {\cos\alpha\cos\beta}.$$
Thus the equation becomes
$$2\sin S\cos D = \frac{2\cos^2S\cos D(1-\cos\alpha\cos\beta)} {\sin S,\cos\alpha\cos\beta}.$$
After cancellation,
$$\sin^2S,\cos\alpha\cos\beta = \cos^2S(1-\cos\alpha\cos\beta).$$
Rearranging gives
$$\cos\alpha\cos\beta=\cos^2S.$$
So the conclusion is correct, but the proof as written does not actually show these steps rigorously.
This is a justification gap, not a fatal logical error, because the missing algebra can be repaired directly.
Step 6: Cancel
$$\cos\frac{\alpha-\beta}{2}$$
after proving it is positive - VALID
The proof correctly checks
$$-\pi<\alpha-\beta<\pi$$
so
$$-\frac\pi2<\frac{\alpha-\beta}{2}<\frac\pi2,$$
hence cosine is strictly positive.
Correct.
Step 7: From
$$\cos\alpha\cos\beta = \cos^2\frac{\alpha+\beta}{2}$$
derive
$$\cos(\alpha-\beta)=1$$
- VALID
Using
$$\cos\alpha\cos\beta = \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}2$$
and
$$\cos^2\frac{\alpha+\beta}{2} = \frac{1+\cos(\alpha+\beta)}2,$$
we obtain
$$\cos(\alpha-\beta)=1.$$
Correct.
Step 8: Deduce $\alpha=\beta$ - VALID
Since
$$-\pi<\alpha-\beta<\pi,$$
the only solution to
$$\cos(\alpha-\beta)=1$$
is
$$\alpha-\beta=0.$$
Correct.
Step 9: Conclude $a=b$ from the sine law - VALID
From
$$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta},$$
and $\alpha=\beta$, we get $a=b$.
Correct.
Completeness Check
The proof handles all nondegenerate triangles.
The positivity needed for cancellation is explicitly justified.
No hidden division by zero occurs:
$$\cos\alpha,\cos\beta$$
could individually vanish in a right triangle, but the proof never divides by them without first introducing expressions where such divisions are already implicit in $\tan\alpha,\tan\beta$. In fact, the original hypothesis itself excludes $\alpha=\pi/2$ or $\beta=\pi/2$, since $\tan\alpha$ or $\tan\beta$ would be undefined. So this issue is harmless.
The main weakness is the central algebraic reduction in Lemma 1. The proof claims a substantial simplification without fully deriving it. The result is correct, but the presentation is incomplete at that point.
The final conclusion does follow from what was proved.
Summary
The overall strategy is correct. The proof uses the sine law to reduce the condition to a trigonometric identity and then derives $\alpha=\beta$.
The only substantial flaw is that the core simplification in Lemma 1 is not rigorously carried out. Several manipulations are sketched, partially abandoned, and then the desired equation is asserted “after cancellation.” A careful grader would require the missing algebra to be written explicitly.
This is a repairable justification gap, not a fundamentally incorrect argument.
VERDICT: FAIL - the crucial reduction in Lemma 1 is asserted without a complete derivation, leaving a nontrivial algebraic gap in the proof.