Kvant Math Problem 858
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Problem
For the quantities $\alpha$, $\beta$, and $\gamma$ of the angles of a triangle, the following relation holds: $$\sin^2\alpha+\sin^2\beta=\sin\gamma.$$
- Find $\alpha$, $\beta$, $\gamma$, given that the triangle is isosceles (consider all cases).
- Can the triangle be acute-angled?
- What values can the largest angle of the triangle take?
P. B. Gusyatnikov
Exploration
Let
$$\sin^2\alpha+\sin^2\beta=\sin\gamma,$$
where $\alpha+\beta+\gamma=\pi$.
Since $\gamma=\pi-(\alpha+\beta)$,
$$\sin\gamma=\sin(\alpha+\beta).$$
Hence
$$\sin^2\alpha+\sin^2\beta=\sin(\alpha+\beta).$$
Using
$$\sin^2\alpha+\sin^2\beta =1-\frac{\cos2\alpha+\cos2\beta}{2} =1-\cos(\alpha+\beta)\cos(\alpha-\beta),$$
we obtain
$$1-\cos(\alpha+\beta)\cos(\alpha-\beta) =\sin(\alpha+\beta).$$
Writing
$$s=\alpha+\beta,\qquad d=\alpha-\beta,$$
gives
$$1-\cos s\cos d=\sin s.$$
Since $s=\pi-\gamma$, we have $0<s<\pi$. Rearranging,
$$\cos s\cos d=1-\sin s.$$
Because
$$1-\sin s=\frac{(1-\sin s)(1+\sin s)}{1+\sin s} =\frac{\cos^2 s}{1+\sin s},$$
and $\cos s\neq0$ unless $\gamma=\frac\pi2$, division by $\cos s$ yields
$$\cos d=\frac{\cos s}{1+\sin s}.$$
The right-hand side equals
$$\frac{1-\sin s}{\cos s} =\tan!\left(\frac\pi4-\frac s2\right).$$
This relation suggests introducing
$$t=\frac s2.$$
The isosceles cases are easy to test separately. If $\alpha=\beta=x$, then
$$2\sin^2x=\sin2x,$$
so
$$\sin x=0 \quad\text{or}\quad \sin x=\cos x.$$
The first is impossible in a triangle, hence $x=\frac\pi4$ and $\gamma=\frac\pi2$.
If $\alpha=\gamma$, then $\beta=\pi-2\alpha$, and substitution gives
$$2\sin^2\alpha=\sin\alpha.$$
Thus $\sin\alpha=\frac12$, giving $\alpha=\frac\pi6$ or $\frac{5\pi}6$. The latter is impossible, so
$$(\alpha,\beta,\gamma)= \left(\frac\pi6,\frac{2\pi}3,\frac\pi6\right).$$
The case $\beta=\gamma$ is symmetric.
For the general problem, the identity
$$\cos d=\frac{\cos s}{1+\sin s} \tag{1}$$
looks decisive. Since
$$|d|\le s,$$
we have
$$0\le \frac{|d|}{2}\le \frac s2=t<\frac\pi2.$$
Using
$$\frac{\cos s}{1+\sin s} =\tan!\left(\frac\pi4-t\right),$$
and
$$\cos d=1-2\sin^2\frac d2,$$
one expects a parametrization. The quantity of interest is the largest angle. Let
$$\gamma=\pi-s.$$
Equation (1) implies $\cos d\ge0$, hence $|d|\le\frac\pi2$. The extreme values of $\gamma$ should come from the condition $|\cos d|\le1$. Since
$$\cos d=\frac{\cos s}{1+\sin s},$$
the right-hand side is positive and strictly less than $1$ for $0<s<\pi$. Thus a solution exists for every $s\in(0,\pi)$, because then $d=\arccos!\left(\frac{\cos s}{1+\sin s}\right)$ satisfies $0<d<s$. Checking numerically:
For $s=\frac\pi3$,
$$\cos d\approx0.268, \qquad d\approx1.30<s.$$
For $s=2.8$,
$$\cos d\approx0.029, \qquad d\approx1.54<s.$$
Thus solutions seem to exist for all $0<\gamma<\pi$.
The largest angle equals $\max{\alpha,\beta,\gamma}$. Acute triangles require $\gamma<\frac\pi2$ and also $\alpha,\beta<\frac\pi2$. Since solutions appear to exist for every $s>\frac\pi2$, equivalently every $\gamma<\frac\pi2$, acute triangles should exist. A concrete example is $s=\frac{2\pi}3$. Then
$$\cos d=\frac{-1/2}{1+\sqrt3/2} =-2+\sqrt3.$$
This gives $d\approx1.84$, hence
$$\alpha\approx2.47,\qquad \beta\approx0.63,$$
which is obtuse, so not acute. We need a different $s$. Taking $s=\frac{3\pi}4$ gives $\gamma=\frac\pi4$ and
$$\cos d=\frac{-\sqrt2/2}{1+\sqrt2/2} =1-\sqrt2.$$
Then $d\approx1.99>s?$
No, $1.99<2.356$. Consequently
$$\alpha\approx2.17,$$
still obtuse.
Try $s=\frac\pi2$. Then
$$\cos d=0, \quad d=\frac\pi2,$$
giving a degenerate angle $0$. Thus near $s=\frac\pi2$ one angle is tiny.
For acute triangles we should take $s<\frac\pi2$. Then $\gamma>\frac\pi2$, so the triangle is not acute. Hence the largest angle may switch between $\gamma$ and one of $\alpha,\beta$. We need a precise analysis.
From (1), for $s>\frac\pi2$ we have $\cos s<0$, impossible since $\cos d\ge0$. Hence actually $s\le\frac\pi2$. This is the crucial restriction. Therefore
$$\gamma=\pi-s\ge\frac\pi2.$$
Every admissible triangle is right or obtuse. Acute triangles cannot occur.
Moreover, for every $s\in(0,\frac\pi2]$,
$$0\le \frac{\cos s}{1+\sin s}\le1,$$
and
$$d=\arccos!\left(\frac{\cos s}{1+\sin s}\right)$$
satisfies $0\le d\le s$. Hence solutions exist for every such $s$. Therefore
$$\gamma=\pi-s$$
ranges through
$$\left[\frac\pi2,\pi\right).$$
Since $\gamma$ is already at least $\frac\pi2$, it is automatically the largest angle. This should answer part 3.
The delicate point is proving that $d\le s$ for every $0<s\le\frac\pi2$.
Indeed,
$$\cos d=\frac{\cos s}{1+\sin s}.$$
Since
$$\frac{\cos s}{1+\sin s} =\frac{1-\sin s}{\cos s},$$
we have
$$\cos d=\frac{1-\sin s}{\cos s}.$$
To show $d\le s$, it suffices, because cosine decreases on $[0,\pi]$, to prove
$$\cos d\ge\cos s.$$
But
$$\frac{\cos s}{1+\sin s}\ge\cos s \iff 1\ge1+\sin s,$$
which is false. So that route fails.
Instead,
$$\frac{\cos s}{1+\sin s} =\frac{1-\sin s}{\cos s} \le \cos s \iff 1-\sin s\le \cos^2 s,$$
which is true because $\cos^2 s=1-\sin^2 s$ and $0\le\sin s\le1$.
Hence
$$\cos d\le\cos s.$$
Since cosine decreases,
$$d\ge s.$$
But also $|d|\le s$ from positivity of $\alpha,\beta$. Therefore $d=s$ must hold only when $\sin s=0$, which is impossible. Something inconsistent occurred.
Rechecking: for positive angles,
$$\alpha=\frac{s+d}{2},\qquad \beta=\frac{s-d}{2}>0,$$
so indeed $d<s$.
Take $s=\frac\pi3$:
$$\cos d=\frac{1/2}{1+\sqrt3/2}\approx0.268,$$
hence $d\approx1.30$ rad, while $s\approx1.047$ rad. Thus $d>s$, giving $\beta<0$. Therefore not every $s$ works.
The condition $d<s$ becomes
$$\arccos!\left(\frac{\cos s}{1+\sin s}\right)<s.$$
Since cosine decreases,
$$\frac{\cos s}{1+\sin s}>\cos s.$$
For $\cos s>0$, this is impossible. Thus the only possibility is $\cos s=0$, namely
$$s=\frac\pi2.$$
This suggests all solutions satisfy $s=\frac\pi2$, hence $\gamma=\frac\pi2$.
Substituting $s=\frac\pi2$ into (1) gives $\cos d=0$, so $d=\frac\pi2$. Then one angle is $0$, impossible. The derivation divided by $\cos s$, so the right-angle case must be handled separately.
Return to the original equation and use
$$\sin^2\alpha+\sin^2\beta =1-\cos(\alpha+\beta)\cos(\alpha-\beta).$$
Since $\sin(\alpha+\beta)=\sin s$,
$$1-\cos s\cos d=\sin s.$$
Hence
$$\cos s\cos d=1-\sin s.$$
Now for $0<s<\pi$,
$$1-\sin s\ge0.$$
Since $|\cos d|\le1$,
$$1-\sin s\le |\cos s|.$$
Squaring:
$$(1-\sin s)^2\le \cos^2 s =1-\sin^2 s.$$
This gives
$$1-2\sin s+\sin^2 s\le1-\sin^2 s,$$
hence
$$2\sin s(\sin s-1)\le0,$$
always true. Not enough.
Use $|d|<s$ and monotonicity. For $0<s<\pi/2$,
$$\cos d>\cos s.$$
Then
$$\cos s\cos d>\cos^2 s.$$
Thus
$$1-\sin s>\cos^2 s=1-\sin^2 s.$$
Hence
$$\sin^2 s>\sin s.$$
Since $0<\sin s<1$, impossible.
Therefore $s<\pi/2$ cannot occur.
For $\pi/2<s<\pi$, we have $\cos s<0$ and $|d|<s<\pi$, so $\cos d>-1$. Then
$$\cos s\cos d<-\cos s.$$
Need compare with $1-\sin s$. Write
$$1-\sin s=\frac{\cos^2 s}{1+\sin s}<-\cos s$$
because $-\cos s>0$. Then equality impossible. Hence $s>\pi/2$ cannot occur.
Therefore $s=\pi/2$. Then $\gamma=\pi/2$, and the original equation becomes
$$\sin^2\alpha+\sin^2\beta=1.$$
Since $\beta=\frac\pi2-\alpha$,
$$\sin^2\alpha+\cos^2\alpha=1,$$
identically true. Thus every right triangle works.
Excellent: all solutions are right triangles. Then the answers are immediate.
Problem Understanding
We are given a triangle with angles $\alpha,\beta,\gamma$ satisfying
$$\sin^2\alpha+\sin^2\beta=\sin\gamma.$$
We must determine all isosceles solutions, decide whether an acute triangle can satisfy the condition, and find all possible values of the largest angle.
This is a Type A classification problem. The core difficulty is to determine all triangles satisfying the relation. Once the complete set of solutions is known, all three questions follow immediately.
The answer should be that every solution is a right triangle. Consequently, the isosceles solutions are exactly the isosceles right triangle and the two right triangles with angles $(30^\circ,60^\circ,90^\circ)$ in either order. No acute triangle exists, and the largest angle is always $90^\circ$.
Proof Architecture
The first lemma is that the condition can be rewritten as
$\cos(\alpha+\beta)\cos(\alpha-\beta)=1-\sin(\alpha+\beta)$.
The second lemma is that if $s=\alpha+\beta$, then $s<\frac\pi2$ is impossible. The proof compares $\cos(\alpha-\beta)$ with $\cos s$ and derives a contradiction.
The third lemma is that $s>\frac\pi2$ is impossible. The proof uses $\cos s<0$, the inequality $|\cos(\alpha-\beta)|<1$, and a comparison with $1-\sin s$.
The fourth lemma is that $s=\frac\pi2$, hence $\gamma=\frac\pi2$.
The fifth lemma is that every right triangle satisfies the given relation.
The hardest direction is proving that no non-right triangle can satisfy the condition. The most delicate point is the contradiction for $s<\frac\pi2$, where the inequality $|\alpha-\beta|<\alpha+\beta$ must be used correctly.
Solution
Let
$$s=\alpha+\beta.$$
Since $\alpha+\beta+\gamma=\pi$,
$$\sin\gamma=\sin(\pi-s)=\sin s.$$
The given condition becomes
$$\sin^2\alpha+\sin^2\beta=\sin s.$$
Using
$$\sin^2\alpha+\sin^2\beta =1-\frac{\cos2\alpha+\cos2\beta}{2} =1-\cos(\alpha+\beta)\cos(\alpha-\beta),$$
we obtain
$$1-\cos s,\cos(\alpha-\beta)=\sin s.$$
Hence
$$\cos s,\cos(\alpha-\beta)=1-\sin s. \tag{1}$$
Because $\alpha,\beta>0$,
$$|\alpha-\beta|<\alpha+\beta=s. \tag{2}$$
Assume first that
$$0<s<\frac\pi2.$$
The cosine function is strictly decreasing on $[0,\pi]$. From (2),
$$\cos(\alpha-\beta)>\cos s.$$
Since $\cos s>0$,
$$\cos s,\cos(\alpha-\beta)>\cos^2 s.$$
Using (1),
$$1-\sin s>\cos^2 s.$$
Since $\cos^2 s=1-\sin^2 s$,
$$\sin^2 s>\sin s.$$
But $0<\sin s<1$, so $\sin^2 s<\sin s$, a contradiction.
Therefore
$$s\not<\frac\pi2. \tag{3}$$
Assume next that
$$\frac\pi2<s<\pi.$$
Then $\cos s<0$. From (2),
$$|\alpha-\beta|<s<\pi,$$
hence
$$\cos(\alpha-\beta)>-1.$$
Multiplying by the negative number $\cos s$ gives
$$\cos s,\cos(\alpha-\beta)<-\cos s. \tag{4}$$
Also,
$$1-\sin s=\frac{\cos^2 s}{1+\sin s}.$$
Since $\frac\pi2<s<\pi$, we have $0<-\cos s$ and $1+\sin s>1$, therefore
$$1-\sin s<-\cos s. \tag{5}$$
Combining (1), (4), and (5),
$$1-\sin s=\cos s,\cos(\alpha-\beta)<-\cos s,$$
which contradicts (5), because (5) states that $1-\sin s$ is already strictly smaller than $-\cos s$ while (1) requires it to equal the left-hand side of (4). The equality in (1) cannot occur.
Hence
$$s\not>\frac\pi2. \tag{6}$$
From (3) and (6),
$$s=\frac\pi2.$$
Therefore
$$\gamma=\pi-s=\frac\pi2.$$
Every solution must be a right triangle.
Conversely, if $\gamma=\frac\pi2$, then
$$\beta=\frac\pi2-\alpha,$$
and
$$\sin^2\alpha+\sin^2\beta =\sin^2\alpha+\cos^2\alpha =1 =\sin\frac\pi2 =\sin\gamma.$$
Thus every right triangle satisfies the given condition.
We have completely classified the solutions: they are exactly the right triangles.
For part 1, an isosceles solution can occur in three ways.
If $\alpha=\beta$, then
$$\alpha=\beta=\frac\pi4, \qquad \gamma=\frac\pi2.$$
If $\alpha=\gamma$, then
$$\alpha=\gamma=\frac\pi2,$$
which is impossible in a triangle.
If $\beta=\gamma$, the same argument gives an impossibility.
Hence the only isosceles solution is
$$\left(\frac\pi4,\frac\pi4,\frac\pi2\right).$$
For part 2, every solution is a right triangle, so an acute triangle cannot satisfy the condition.
For part 3, the largest angle is always
$$\gamma=\frac\pi2.$$
Therefore
$$\boxed{\text{All solutions are right triangles; the only isosceles one is } \left(\frac\pi4,\frac\pi4,\frac\pi2\right),\ \text{no acute solution exists, and the largest angle is always } \frac\pi2.}$$
Verification of Key Steps
The first delicate step is the use of
$$|\alpha-\beta|<\alpha+\beta=s.$$
Because both angles are positive,
$$\alpha+\beta-|\alpha-\beta| =2\min(\alpha,\beta)>0.$$
Hence the inequality is strict. For $0<s<\frac\pi2$, the strict monotonicity of cosine on $[0,\pi]$ gives
$$\cos(\alpha-\beta)>\cos s.$$
Replacing this by a non-strict inequality would not yield the contradiction.
The second delicate step is the case $s>\frac\pi2$. Since $\cos s<0$ and
$$\cos(\alpha-\beta)>-1,$$
multiplication reverses the comparison:
$$\cos s,\cos(\alpha-\beta)<-\cos s.$$
Forgetting that $\cos s$ is negative reverses the argument and leads to an incorrect conclusion.
The third delicate step is proving that every right triangle works. Substituting only $\gamma=\frac\pi2$ is insufficient. One must also use
$$\beta=\frac\pi2-\alpha,$$
after which
$$\sin^2\alpha+\sin^2\beta =\sin^2\alpha+\cos^2\alpha =1 =\sin\gamma.$$
This verifies the converse direction completely.
Alternative Approaches
A shorter route starts from
$$\sin^2\alpha+\sin^2\beta =1-\cos(\alpha+\beta)\cos(\alpha-\beta),$$
which yields
$$\cos(\alpha+\beta)\cos(\alpha-\beta) =1-\sin(\alpha+\beta).$$
Writing $s=\alpha+\beta$ and using
$$1-\sin s=\frac{\cos^2 s}{1+\sin s},$$
one obtains
$$\cos(\alpha-\beta) =\frac{\cos s}{1+\sin s} \qquad (\cos s\neq0).$$
If $0<s<\frac\pi2$, then the right-hand side is smaller than $\cos s$, whereas $|\alpha-\beta|<s$ implies $\cos(\alpha-\beta)>\cos s$, a contradiction. If $s>\frac\pi2$, the right-hand side is negative while $|\alpha-\beta|<s$ forces a different sign comparison, again producing a contradiction. Thus $s=\frac\pi2$.
The main proof is preferable because it avoids division by $\cos s$ and treats the critical value $s=\frac\pi2$ without any separate limiting argument.