Distance between incentre and orthocentre.area of a triangle formulaProve that the triangles $ABC$ and $AB^'C^'$ have the same incentre.Olympiad Trigonometric InequalityBrocard Angles proof by Sine and cosine formulae.Coordinates of orthocentre,circumcentre and incentre of a triangle formed in 3d planeProve $r^2+r_1^2+r_2^2+r_3^2=16R^2-a^2-b^2-c^2$ for a triangle with sides $a$, $b$, $c$, circumradius $R$, inradius $r$, exradii $r_1$, $r_2$, $r_3$A question on an isosceles obtuse angled triangle.What is the angle between the median and the bisector?Sides of a triangle given perimeter and two anglesDistance between incentre and excentre of a triangle
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Distance between incentre and orthocentre.
area of a triangle formulaProve that the triangles $ABC$ and $AB^'C^'$ have the same incentre.Olympiad Trigonometric InequalityBrocard Angles proof by Sine and cosine formulae.Coordinates of orthocentre,circumcentre and incentre of a triangle formed in 3d planeProve $r^2+r_1^2+r_2^2+r_3^2=16R^2-a^2-b^2-c^2$ for a triangle with sides $a$, $b$, $c$, circumradius $R$, inradius $r$, exradii $r_1$, $r_2$, $r_3$A question on an isosceles obtuse angled triangle.What is the angle between the median and the bisector?Sides of a triangle given perimeter and two anglesDistance between incentre and excentre of a triangle
$begingroup$
I want to prove that the distance between incentre and orthocentre is $$sqrt2r^2-4R^2cos Acos Bcos C $$here $r$ is inradius and $R$ is circumradius.
I considered $triangle API$ ($P$ is orthocentre and $I$ is incentre). I could find $AP=2Rcos A$, $AI=4RsinfracB2sinfracC2 $ and $angle PAI=angle fracB-C2$.
So applying cosine rule I got $$PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
How to proceed form here ?
geometry trigonometry euclidean-geometry triangle
edited May 2 '16 at 5:09
Michael Hardy
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
$endgroup$
add a comment |
$begingroup$
I want to prove that the distance between incentre and orthocentre is $$sqrt2r^2-4R^2cos Acos Bcos C $$here $r$ is inradius and $R$ is circumradius.
I considered $triangle API$ ($P$ is orthocentre and $I$ is incentre). I could find $AP=2Rcos A$, $AI=4RsinfracB2sinfracC2 $ and $angle PAI=angle fracB-C2$.
So applying cosine rule I got $$PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
How to proceed form here ?
geometry trigonometry euclidean-geometry triangle
edited May 2 '16 at 5:09
Michael Hardy
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
$endgroup$
add a comment |
$begingroup$
I want to prove that the distance between incentre and orthocentre is $$sqrt2r^2-4R^2cos Acos Bcos C $$here $r$ is inradius and $R$ is circumradius.
I considered $triangle API$ ($P$ is orthocentre and $I$ is incentre). I could find $AP=2Rcos A$, $AI=4RsinfracB2sinfracC2 $ and $angle PAI=angle fracB-C2$.
So applying cosine rule I got $$PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
How to proceed form here ?
geometry trigonometry euclidean-geometry triangle
edited May 2 '16 at 5:09
Michael Hardy
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
$endgroup$
I want to prove that the distance between incentre and orthocentre is $$sqrt2r^2-4R^2cos Acos Bcos C $$here $r$ is inradius and $R$ is circumradius.
I considered $triangle API$ ($P$ is orthocentre and $I$ is incentre). I could find $AP=2Rcos A$, $AI=4RsinfracB2sinfracC2 $ and $angle PAI=angle fracB-C2$.
So applying cosine rule I got $$PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
How to proceed form here ?
geometry trigonometry euclidean-geometry triangle
geometry trigonometry euclidean-geometry triangle
edited May 2 '16 at 5:09
Michael Hardy
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
edited May 2 '16 at 5:09
Michael Hardy
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
edited May 2 '16 at 5:09
Michael Hardy
1
edited May 2 '16 at 5:09
Michael Hardy
1
edited May 2 '16 at 5:09
Michael Hardy
1
1
asked May 2 '16 at 5:07
mathemathermathemather
864622
asked May 2 '16 at 5:07
mathemathermathemather
864622
asked May 2 '16 at 5:07
mathemathermathemather
864622
864622
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So applying cosine rule I got $$small PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
I think that you have a typo (the red part) :
$$beginalign&small PI^2=4R^2colorredcos^2A+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2 -4cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)right)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2-cos Acdot colorgreen2sinfracB2cosfracB2cdot 2sinfracC2cosfracC2-4cos Asin^2fracB2sin^2fracC2right)\&small=4R^2left(cos^2A+8sin^2fracB2sin^2fracC2cdot colorbluefrac 12left(1-cos Aright)-cos Acolorgreensin Bsin Cright)\&small=4R^2left(cos^2A+8colorbluesin^2frac A2sin^2fracB2sin^2fracC2-cos Asin Bsin Cright)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos A-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos(180^circ-(B+C))-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2-cos Acos Bcos Cright)\&small =2left(4Rsin frac A2sin frac B2sinfrac C2right)^2-4R^2cos Acos Bcos C\&small=2r^2-4R^2cos Acos Bcos Cendalign$$
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
$endgroup$
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
add a comment |
$begingroup$
Use these facts
1. The Euler circle is tangent to the inscribed circle
2. The distance between the circumcenter and the incenter using the Euler formula.
3. The formula for the power of a point with respect to a circle
4. The properties of the Euler line
5. The fact that the reflection of the orthocenter with respect to any side of a triangle is on the circumcircle
6. the relationship between the median, the two adjacent sides to that median and the third side.
Then just do the algebra
Let O be the circumcenter (X(3), H the orthocenter (X(4)),I the incenter (X(1)), and W The center of the Euler circle (X(5)), and A' the foot of the altitude on the corresponding side. Assuming a triangle ABC
We have OI^2 =R^2 -2Rr where R is the circumradius and r the inscribed circle radius(
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So applying cosine rule I got $$small PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
I think that you have a typo (the red part) :
$$beginalign&small PI^2=4R^2colorredcos^2A+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2 -4cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)right)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2-cos Acdot colorgreen2sinfracB2cosfracB2cdot 2sinfracC2cosfracC2-4cos Asin^2fracB2sin^2fracC2right)\&small=4R^2left(cos^2A+8sin^2fracB2sin^2fracC2cdot colorbluefrac 12left(1-cos Aright)-cos Acolorgreensin Bsin Cright)\&small=4R^2left(cos^2A+8colorbluesin^2frac A2sin^2fracB2sin^2fracC2-cos Asin Bsin Cright)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos A-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos(180^circ-(B+C))-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2-cos Acos Bcos Cright)\&small =2left(4Rsin frac A2sin frac B2sinfrac C2right)^2-4R^2cos Acos Bcos C\&small=2r^2-4R^2cos Acos Bcos Cendalign$$
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
$endgroup$
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
add a comment |
$begingroup$
So applying cosine rule I got $$small PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
I think that you have a typo (the red part) :
$$beginalign&small PI^2=4R^2colorredcos^2A+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2 -4cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)right)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2-cos Acdot colorgreen2sinfracB2cosfracB2cdot 2sinfracC2cosfracC2-4cos Asin^2fracB2sin^2fracC2right)\&small=4R^2left(cos^2A+8sin^2fracB2sin^2fracC2cdot colorbluefrac 12left(1-cos Aright)-cos Acolorgreensin Bsin Cright)\&small=4R^2left(cos^2A+8colorbluesin^2frac A2sin^2fracB2sin^2fracC2-cos Asin Bsin Cright)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos A-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos(180^circ-(B+C))-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2-cos Acos Bcos Cright)\&small =2left(4Rsin frac A2sin frac B2sinfrac C2right)^2-4R^2cos Acos Bcos C\&small=2r^2-4R^2cos Acos Bcos Cendalign$$
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
$endgroup$
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
add a comment |
$begingroup$
So applying cosine rule I got $$small PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
I think that you have a typo (the red part) :
$$beginalign&small PI^2=4R^2colorredcos^2A+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2 -4cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)right)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2-cos Acdot colorgreen2sinfracB2cosfracB2cdot 2sinfracC2cosfracC2-4cos Asin^2fracB2sin^2fracC2right)\&small=4R^2left(cos^2A+8sin^2fracB2sin^2fracC2cdot colorbluefrac 12left(1-cos Aright)-cos Acolorgreensin Bsin Cright)\&small=4R^2left(cos^2A+8colorbluesin^2frac A2sin^2fracB2sin^2fracC2-cos Asin Bsin Cright)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos A-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos(180^circ-(B+C))-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2-cos Acos Bcos Cright)\&small =2left(4Rsin frac A2sin frac B2sinfrac C2right)^2-4R^2cos Acos Bcos C\&small=2r^2-4R^2cos Acos Bcos Cendalign$$
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
$endgroup$
So applying cosine rule I got $$small PI^2=4R^2+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)$$
I think that you have a typo (the red part) :
$$beginalign&small PI^2=4R^2colorredcos^2A+16R^2sin^2fracB2sin^2fracC2 -16R^2cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2 -4cos AsinfracB2sinfracC2Bigg(cosfracB2cosfracC2+sinfracB2sinfracC2Bigg)right)\&small=4R^2left(cos^2A+4sin^2fracB2sin^2fracC2-cos Acdot colorgreen2sinfracB2cosfracB2cdot 2sinfracC2cosfracC2-4cos Asin^2fracB2sin^2fracC2right)\&small=4R^2left(cos^2A+8sin^2fracB2sin^2fracC2cdot colorbluefrac 12left(1-cos Aright)-cos Acolorgreensin Bsin Cright)\&small=4R^2left(cos^2A+8colorbluesin^2frac A2sin^2fracB2sin^2fracC2-cos Asin Bsin Cright)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos A-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2+cos A(cos(180^circ-(B+C))-sin Bsin C)right)\&small=4R^2left(8sin^2frac A2sin^2fracB2sin^2fracC2-cos Acos Bcos Cright)\&small =2left(4Rsin frac A2sin frac B2sinfrac C2right)^2-4R^2cos Acos Bcos C\&small=2r^2-4R^2cos Acos Bcos Cendalign$$
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
answered May 2 '16 at 6:38
mathlovemathlove
91.8k882218
91.8k882218
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
add a comment |
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
Thank you. The use of different colours really made it easy for me to understand.
$endgroup$
– mathemather
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
$begingroup$
@arutoregni: You are welcome.
$endgroup$
– mathlove
May 2 '16 at 8:15
add a comment |
$begingroup$
Use these facts
1. The Euler circle is tangent to the inscribed circle
2. The distance between the circumcenter and the incenter using the Euler formula.
3. The formula for the power of a point with respect to a circle
4. The properties of the Euler line
5. The fact that the reflection of the orthocenter with respect to any side of a triangle is on the circumcircle
6. the relationship between the median, the two adjacent sides to that median and the third side.
Then just do the algebra
Let O be the circumcenter (X(3), H the orthocenter (X(4)),I the incenter (X(1)), and W The center of the Euler circle (X(5)), and A' the foot of the altitude on the corresponding side. Assuming a triangle ABC
We have OI^2 =R^2 -2Rr where R is the circumradius and r the inscribed circle radius(
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Use these facts
1. The Euler circle is tangent to the inscribed circle
2. The distance between the circumcenter and the incenter using the Euler formula.
3. The formula for the power of a point with respect to a circle
4. The properties of the Euler line
5. The fact that the reflection of the orthocenter with respect to any side of a triangle is on the circumcircle
6. the relationship between the median, the two adjacent sides to that median and the third side.
Then just do the algebra
Let O be the circumcenter (X(3), H the orthocenter (X(4)),I the incenter (X(1)), and W The center of the Euler circle (X(5)), and A' the foot of the altitude on the corresponding side. Assuming a triangle ABC
We have OI^2 =R^2 -2Rr where R is the circumradius and r the inscribed circle radius(
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Use these facts
1. The Euler circle is tangent to the inscribed circle
2. The distance between the circumcenter and the incenter using the Euler formula.
3. The formula for the power of a point with respect to a circle
4. The properties of the Euler line
5. The fact that the reflection of the orthocenter with respect to any side of a triangle is on the circumcircle
6. the relationship between the median, the two adjacent sides to that median and the third side.
Then just do the algebra
Let O be the circumcenter (X(3), H the orthocenter (X(4)),I the incenter (X(1)), and W The center of the Euler circle (X(5)), and A' the foot of the altitude on the corresponding side. Assuming a triangle ABC
We have OI^2 =R^2 -2Rr where R is the circumradius and r the inscribed circle radius(
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Use these facts
1. The Euler circle is tangent to the inscribed circle
2. The distance between the circumcenter and the incenter using the Euler formula.
3. The formula for the power of a point with respect to a circle
4. The properties of the Euler line
5. The fact that the reflection of the orthocenter with respect to any side of a triangle is on the circumcircle
6. the relationship between the median, the two adjacent sides to that median and the third side.
Then just do the algebra
Let O be the circumcenter (X(3), H the orthocenter (X(4)),I the incenter (X(1)), and W The center of the Euler circle (X(5)), and A' the foot of the altitude on the corresponding side. Assuming a triangle ABC
We have OI^2 =R^2 -2Rr where R is the circumradius and r the inscribed circle radius(
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
claudeclaude
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
claudeclaude
1
answered yesterday
claudeclaude
1
1
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
claude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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