Mixing
Inequalities among the slopes of the sides and the slopes of internal angle bisectors
$begingroup$
Suppose a triangle ABC without sides nor internal angle bisectors parallel to y-axis, and let $m_1, m_2, m_3 $ be the slopes of the equations of sides BC, CA, AB, and $s_1, s_2, s_3$ the slopes of the equations of internal bisectors of angles A, B, C (the cartesian axes being orthogonal or oblique). Having made these definitions, the following inequalities hold true:
$$(m_1-m_2)(m_2-s_1)(s_1-s_2)(s_2-m_1)<0$$
$$(m_2-m_3)(m_3-s_2)(s_2-s_3)(s_3-m_2)<0$$
$$(m_3-m_1)(m_1-s_3)(s_3-s_1)(s_1-m_3)<0$$
In order to prove these inequalities one needs only to know the concave quadrilateral slope inequality (The concave quadrilateral and the slopes of its sides) and note that the quadrilaterals BCAI, CABI, ABCI, are concave (I being the incenter of ABC).
Another curious inequalities, perhaps never mentioned in the mathematical literature, are these, which hold true if the cartesian axes are orthogonal:
$$(1+m_1s_2)(1+m_1s_3)(1+ s_2s_3)>0$$
$$(1+m_2s_3)(1+m_2s_1)(1+ s_3s_1)>0$$
$$(1+m_3s_1)(1+m_3s_2)(1+ s_1s_2)>0$$
They can be easily proved if one knows the obtuse angled triangle slope inequality (A theorem for classifying triangles when given only the slopes of the equations of its sides) and note that triangles BCI, CAI, ABI, are obtuse angled.
Does anyone know different proofs for these inequalities?
geometry inequality analytic-geometry triangles alternative-proof
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
$endgroup$
add a comment |
$begingroup$
Suppose a triangle ABC without sides nor internal angle bisectors parallel to y-axis, and let $m_1, m_2, m_3 $ be the slopes of the equations of sides BC, CA, AB, and $s_1, s_2, s_3$ the slopes of the equations of internal bisectors of angles A, B, C (the cartesian axes being orthogonal or oblique). Having made these definitions, the following inequalities hold true:
$$(m_1-m_2)(m_2-s_1)(s_1-s_2)(s_2-m_1)<0$$
$$(m_2-m_3)(m_3-s_2)(s_2-s_3)(s_3-m_2)<0$$
$$(m_3-m_1)(m_1-s_3)(s_3-s_1)(s_1-m_3)<0$$
In order to prove these inequalities one needs only to know the concave quadrilateral slope inequality (The concave quadrilateral and the slopes of its sides) and note that the quadrilaterals BCAI, CABI, ABCI, are concave (I being the incenter of ABC).
Another curious inequalities, perhaps never mentioned in the mathematical literature, are these, which hold true if the cartesian axes are orthogonal:
$$(1+m_1s_2)(1+m_1s_3)(1+ s_2s_3)>0$$
$$(1+m_2s_3)(1+m_2s_1)(1+ s_3s_1)>0$$
$$(1+m_3s_1)(1+m_3s_2)(1+ s_1s_2)>0$$
They can be easily proved if one knows the obtuse angled triangle slope inequality (A theorem for classifying triangles when given only the slopes of the equations of its sides) and note that triangles BCI, CAI, ABI, are obtuse angled.
Does anyone know different proofs for these inequalities?
geometry inequality analytic-geometry triangles alternative-proof
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
$endgroup$
$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50
add a comment |
$begingroup$
Suppose a triangle ABC without sides nor internal angle bisectors parallel to y-axis, and let $m_1, m_2, m_3 $ be the slopes of the equations of sides BC, CA, AB, and $s_1, s_2, s_3$ the slopes of the equations of internal bisectors of angles A, B, C (the cartesian axes being orthogonal or oblique). Having made these definitions, the following inequalities hold true:
$$(m_1-m_2)(m_2-s_1)(s_1-s_2)(s_2-m_1)<0$$
$$(m_2-m_3)(m_3-s_2)(s_2-s_3)(s_3-m_2)<0$$
$$(m_3-m_1)(m_1-s_3)(s_3-s_1)(s_1-m_3)<0$$
In order to prove these inequalities one needs only to know the concave quadrilateral slope inequality (The concave quadrilateral and the slopes of its sides) and note that the quadrilaterals BCAI, CABI, ABCI, are concave (I being the incenter of ABC).
Another curious inequalities, perhaps never mentioned in the mathematical literature, are these, which hold true if the cartesian axes are orthogonal:
$$(1+m_1s_2)(1+m_1s_3)(1+ s_2s_3)>0$$
$$(1+m_2s_3)(1+m_2s_1)(1+ s_3s_1)>0$$
$$(1+m_3s_1)(1+m_3s_2)(1+ s_1s_2)>0$$
They can be easily proved if one knows the obtuse angled triangle slope inequality (A theorem for classifying triangles when given only the slopes of the equations of its sides) and note that triangles BCI, CAI, ABI, are obtuse angled.
Does anyone know different proofs for these inequalities?
geometry inequality analytic-geometry triangles alternative-proof
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
$endgroup$
Suppose a triangle ABC without sides nor internal angle bisectors parallel to y-axis, and let $m_1, m_2, m_3 $ be the slopes of the equations of sides BC, CA, AB, and $s_1, s_2, s_3$ the slopes of the equations of internal bisectors of angles A, B, C (the cartesian axes being orthogonal or oblique). Having made these definitions, the following inequalities hold true:
$$(m_1-m_2)(m_2-s_1)(s_1-s_2)(s_2-m_1)<0$$
$$(m_2-m_3)(m_3-s_2)(s_2-s_3)(s_3-m_2)<0$$
$$(m_3-m_1)(m_1-s_3)(s_3-s_1)(s_1-m_3)<0$$
In order to prove these inequalities one needs only to know the concave quadrilateral slope inequality (The concave quadrilateral and the slopes of its sides) and note that the quadrilaterals BCAI, CABI, ABCI, are concave (I being the incenter of ABC).
Another curious inequalities, perhaps never mentioned in the mathematical literature, are these, which hold true if the cartesian axes are orthogonal:
$$(1+m_1s_2)(1+m_1s_3)(1+ s_2s_3)>0$$
$$(1+m_2s_3)(1+m_2s_1)(1+ s_3s_1)>0$$
$$(1+m_3s_1)(1+m_3s_2)(1+ s_1s_2)>0$$
They can be easily proved if one knows the obtuse angled triangle slope inequality (A theorem for classifying triangles when given only the slopes of the equations of its sides) and note that triangles BCI, CAI, ABI, are obtuse angled.
Does anyone know different proofs for these inequalities?
geometry inequality analytic-geometry triangles alternative-proof
geometry inequality analytic-geometry triangles alternative-proof
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
edited Oct 26 '18 at 16:26
MrDudulex
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
asked Oct 26 '18 at 16:08
MrDudulexMrDudulex
439111
439111
$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50
add a comment |
$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50
$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50
$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50
add a comment |
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$begingroup$
The first inequalities are immediate when you consider that each factor amounts to "sine of the oriented angle between two lines, divided by two cosines". Each denominator cosine appears twice across the full product, so contributes nothing to the sign. On the other hand, one of the factors has a different orientation from the other three, so that the sign of the product is negative. The second set of inequalities reduces to a product of cosines. Perhaps your investigations would benefit by thinking less in terms of slopes and more in terms of directions (that is, direction vectors).
$endgroup$
– Blue
Oct 26 '18 at 18:50