Mixing
In the triangle $triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $angle ACB > 60^circ$.
$begingroup$
The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $angle ACB > 60^circ$.
In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $alpha = angle BAC, beta = angle CBA, gamma = angle ACB$. Suppose $gamma le 60^circ$:
$$gamma le 60^circ iff sin(gamma) leq frac{sqrt{3}}{2}$$
Applying the sine rule:
$$frac{sin(alpha)}{|BC|} = frac{sin(beta)}{|AC|} = frac{sin(gamma)}{|AB|}$$
$$ifffrac{|AC|^3}{|AB|^3} + frac{|BC|^3}{|AB|^3} = frac{sin^3(alpha) + sin^3(beta)}{sin^3(gamma)} = 1$$
$$iff sin^3(alpha) + sin^3(beta) = sin^3(gamma) le left( frac{sqrt{3}}{2} right)^3$$
$$iffsin(alpha) le frac{sqrt(3)}{2}, , sin(beta) le frac{sqrt(3)}{2}tag{1}$$
We also know that:
$$alpha + beta = 180^circ - gamma ge 120^circtag{2}$$
$$alpha + beta < 180^circtag{3}$$
Without loss of generality, assume $alpha ge beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a ge 120^circ, , beta le 60^circ$$
$$implies |BC| > |AB| qquad unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?
geometry proof-verification proof-writing contest-math alternative-proof
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $angle ACB > 60^circ$.
In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $alpha = angle BAC, beta = angle CBA, gamma = angle ACB$. Suppose $gamma le 60^circ$:
$$gamma le 60^circ iff sin(gamma) leq frac{sqrt{3}}{2}$$
Applying the sine rule:
$$frac{sin(alpha)}{|BC|} = frac{sin(beta)}{|AC|} = frac{sin(gamma)}{|AB|}$$
$$ifffrac{|AC|^3}{|AB|^3} + frac{|BC|^3}{|AB|^3} = frac{sin^3(alpha) + sin^3(beta)}{sin^3(gamma)} = 1$$
$$iff sin^3(alpha) + sin^3(beta) = sin^3(gamma) le left( frac{sqrt{3}}{2} right)^3$$
$$iffsin(alpha) le frac{sqrt(3)}{2}, , sin(beta) le frac{sqrt(3)}{2}tag{1}$$
We also know that:
$$alpha + beta = 180^circ - gamma ge 120^circtag{2}$$
$$alpha + beta < 180^circtag{3}$$
Without loss of generality, assume $alpha ge beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a ge 120^circ, , beta le 60^circ$$
$$implies |BC| > |AB| qquad unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?
geometry proof-verification proof-writing contest-math alternative-proof
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
$endgroup$
4
$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59
add a comment |
$begingroup$
The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $angle ACB > 60^circ$.
In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $alpha = angle BAC, beta = angle CBA, gamma = angle ACB$. Suppose $gamma le 60^circ$:
$$gamma le 60^circ iff sin(gamma) leq frac{sqrt{3}}{2}$$
Applying the sine rule:
$$frac{sin(alpha)}{|BC|} = frac{sin(beta)}{|AC|} = frac{sin(gamma)}{|AB|}$$
$$ifffrac{|AC|^3}{|AB|^3} + frac{|BC|^3}{|AB|^3} = frac{sin^3(alpha) + sin^3(beta)}{sin^3(gamma)} = 1$$
$$iff sin^3(alpha) + sin^3(beta) = sin^3(gamma) le left( frac{sqrt{3}}{2} right)^3$$
$$iffsin(alpha) le frac{sqrt(3)}{2}, , sin(beta) le frac{sqrt(3)}{2}tag{1}$$
We also know that:
$$alpha + beta = 180^circ - gamma ge 120^circtag{2}$$
$$alpha + beta < 180^circtag{3}$$
Without loss of generality, assume $alpha ge beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a ge 120^circ, , beta le 60^circ$$
$$implies |BC| > |AB| qquad unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?
geometry proof-verification proof-writing contest-math alternative-proof
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
$endgroup$
The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $angle ACB > 60^circ$.
In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $alpha = angle BAC, beta = angle CBA, gamma = angle ACB$. Suppose $gamma le 60^circ$:
$$gamma le 60^circ iff sin(gamma) leq frac{sqrt{3}}{2}$$
Applying the sine rule:
$$frac{sin(alpha)}{|BC|} = frac{sin(beta)}{|AC|} = frac{sin(gamma)}{|AB|}$$
$$ifffrac{|AC|^3}{|AB|^3} + frac{|BC|^3}{|AB|^3} = frac{sin^3(alpha) + sin^3(beta)}{sin^3(gamma)} = 1$$
$$iff sin^3(alpha) + sin^3(beta) = sin^3(gamma) le left( frac{sqrt{3}}{2} right)^3$$
$$iffsin(alpha) le frac{sqrt(3)}{2}, , sin(beta) le frac{sqrt(3)}{2}tag{1}$$
We also know that:
$$alpha + beta = 180^circ - gamma ge 120^circtag{2}$$
$$alpha + beta < 180^circtag{3}$$
Without loss of generality, assume $alpha ge beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a ge 120^circ, , beta le 60^circ$$
$$implies |BC| > |AB| qquad unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?
geometry proof-verification proof-writing contest-math alternative-proof
geometry proof-verification proof-writing contest-math alternative-proof
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
edited Feb 2 at 7:17
Michael Rozenberg
110k1896201
110k1896201
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
asked Feb 1 at 23:44
jvdhooftjvdhooft
5,65961641
5,65961641
4
$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59
add a comment |
4
$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59
4
4
$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59
$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In the standard notation we need to prove that
$$frac{a^2+b^2-c^2}{2ab}<cos60^{circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$measuredangle ACB>60^{circ}$$ and we are done!
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
With your answer, to justify $alpha ge 120°$ means that you need to exclude the possibility of $alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.
Also, here is an alternative solution. With the equation of
$$leftlvert AB rightrvert^3 = leftlvert AC rightrvert^3 + leftlvert BC rightrvert^3 tag{1}label{eq1}$$
since all lengths are positive values, you automatically have that
$$leftlvert AB rightrvert gt leftlvert AC rightrvert tag{2}label{eq2}$$
$$leftlvert AB rightrvert gt leftlvert BC rightrvert tag{3}label{eq3}$$
Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $angle ACB le 60°$, then the other $2$ angles are $lt 60°$, giving a sum $lt 180°$, which is not true. As such, $angle ACB gt 60°$.
Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
$endgroup$
add a comment |
$begingroup$
$blacksquare$
Alternatively, there's a simpler way to prove that indeed $angle ACB>60º$.
Lemma (Proposition 18 from Euclid's Elements)
In any triangle the angle opposite the greater side is greater
From $|AB|^3=|AC|^3+|BC|^3$ we deduce
$|AB|>|AC| ; text{and} ; |AB|>|BC|$
It follows that $$angle ACB>angle BAC ; text{and} ;angle ACB>angle CBA$$
Therefore $$angle ACB>frac{180°}{3}=60°$$
since the triangle isn't equilateral. $;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;square$
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the standard notation we need to prove that
$$frac{a^2+b^2-c^2}{2ab}<cos60^{circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$measuredangle ACB>60^{circ}$$ and we are done!
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
In the standard notation we need to prove that
$$frac{a^2+b^2-c^2}{2ab}<cos60^{circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$measuredangle ACB>60^{circ}$$ and we are done!
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
In the standard notation we need to prove that
$$frac{a^2+b^2-c^2}{2ab}<cos60^{circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$measuredangle ACB>60^{circ}$$ and we are done!
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
In the standard notation we need to prove that
$$frac{a^2+b^2-c^2}{2ab}<cos60^{circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$measuredangle ACB>60^{circ}$$ and we are done!
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 2 at 7:17
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
With your answer, to justify $alpha ge 120°$ means that you need to exclude the possibility of $alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.
Also, here is an alternative solution. With the equation of
$$leftlvert AB rightrvert^3 = leftlvert AC rightrvert^3 + leftlvert BC rightrvert^3 tag{1}label{eq1}$$
since all lengths are positive values, you automatically have that
$$leftlvert AB rightrvert gt leftlvert AC rightrvert tag{2}label{eq2}$$
$$leftlvert AB rightrvert gt leftlvert BC rightrvert tag{3}label{eq3}$$
Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $angle ACB le 60°$, then the other $2$ angles are $lt 60°$, giving a sum $lt 180°$, which is not true. As such, $angle ACB gt 60°$.
Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
$endgroup$
add a comment |
$begingroup$
With your answer, to justify $alpha ge 120°$ means that you need to exclude the possibility of $alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.
Also, here is an alternative solution. With the equation of
$$leftlvert AB rightrvert^3 = leftlvert AC rightrvert^3 + leftlvert BC rightrvert^3 tag{1}label{eq1}$$
since all lengths are positive values, you automatically have that
$$leftlvert AB rightrvert gt leftlvert AC rightrvert tag{2}label{eq2}$$
$$leftlvert AB rightrvert gt leftlvert BC rightrvert tag{3}label{eq3}$$
Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $angle ACB le 60°$, then the other $2$ angles are $lt 60°$, giving a sum $lt 180°$, which is not true. As such, $angle ACB gt 60°$.
Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
$endgroup$
add a comment |
$begingroup$
With your answer, to justify $alpha ge 120°$ means that you need to exclude the possibility of $alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.
Also, here is an alternative solution. With the equation of
$$leftlvert AB rightrvert^3 = leftlvert AC rightrvert^3 + leftlvert BC rightrvert^3 tag{1}label{eq1}$$
since all lengths are positive values, you automatically have that
$$leftlvert AB rightrvert gt leftlvert AC rightrvert tag{2}label{eq2}$$
$$leftlvert AB rightrvert gt leftlvert BC rightrvert tag{3}label{eq3}$$
Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $angle ACB le 60°$, then the other $2$ angles are $lt 60°$, giving a sum $lt 180°$, which is not true. As such, $angle ACB gt 60°$.
Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
$endgroup$
With your answer, to justify $alpha ge 120°$ means that you need to exclude the possibility of $alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.
Also, here is an alternative solution. With the equation of
$$leftlvert AB rightrvert^3 = leftlvert AC rightrvert^3 + leftlvert BC rightrvert^3 tag{1}label{eq1}$$
since all lengths are positive values, you automatically have that
$$leftlvert AB rightrvert gt leftlvert AC rightrvert tag{2}label{eq2}$$
$$leftlvert AB rightrvert gt leftlvert BC rightrvert tag{3}label{eq3}$$
Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $angle ACB le 60°$, then the other $2$ angles are $lt 60°$, giving a sum $lt 180°$, which is not true. As such, $angle ACB gt 60°$.
Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
edited Feb 2 at 2:02
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
answered Feb 2 at 0:37
John OmielanJohn Omielan
4,9712217
4,9712217
add a comment |
add a comment |
$begingroup$
$blacksquare$
Alternatively, there's a simpler way to prove that indeed $angle ACB>60º$.
Lemma (Proposition 18 from Euclid's Elements)
In any triangle the angle opposite the greater side is greater
From $|AB|^3=|AC|^3+|BC|^3$ we deduce
$|AB|>|AC| ; text{and} ; |AB|>|BC|$
It follows that $$angle ACB>angle BAC ; text{and} ;angle ACB>angle CBA$$
Therefore $$angle ACB>frac{180°}{3}=60°$$
since the triangle isn't equilateral. $;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;square$
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
$begingroup$
$blacksquare$
Alternatively, there's a simpler way to prove that indeed $angle ACB>60º$.
Lemma (Proposition 18 from Euclid's Elements)
In any triangle the angle opposite the greater side is greater
From $|AB|^3=|AC|^3+|BC|^3$ we deduce
$|AB|>|AC| ; text{and} ; |AB|>|BC|$
It follows that $$angle ACB>angle BAC ; text{and} ;angle ACB>angle CBA$$
Therefore $$angle ACB>frac{180°}{3}=60°$$
since the triangle isn't equilateral. $;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;square$
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
$begingroup$
$blacksquare$
Alternatively, there's a simpler way to prove that indeed $angle ACB>60º$.
Lemma (Proposition 18 from Euclid's Elements)
In any triangle the angle opposite the greater side is greater
From $|AB|^3=|AC|^3+|BC|^3$ we deduce
$|AB|>|AC| ; text{and} ; |AB|>|BC|$
It follows that $$angle ACB>angle BAC ; text{and} ;angle ACB>angle CBA$$
Therefore $$angle ACB>frac{180°}{3}=60°$$
since the triangle isn't equilateral. $;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;square$
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
$endgroup$
$blacksquare$
Alternatively, there's a simpler way to prove that indeed $angle ACB>60º$.
Lemma (Proposition 18 from Euclid's Elements)
In any triangle the angle opposite the greater side is greater
From $|AB|^3=|AC|^3+|BC|^3$ we deduce
$|AB|>|AC| ; text{and} ; |AB|>|BC|$
It follows that $$angle ACB>angle BAC ; text{and} ;angle ACB>angle CBA$$
Therefore $$angle ACB>frac{180°}{3}=60°$$
since the triangle isn't equilateral. $;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;square$
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
answered Feb 2 at 16:19
Dr. MathvaDr. Mathva
3,493630
3,493630
add a comment |
add a comment |
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$begingroup$
You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $angle ACB $ is the largest angle.
$endgroup$
– WW1
Feb 1 at 23:59