A problem related to circle , altitude , triangle. The 2019 Stack Overflow Developer Survey Results Are InFinding value of an angle in a triangle.In △ABC, median AM = 17, altitude AD = 15 and the circum-radius R = 10. Find BC^2A circle is inscribed in sector of another bigger circle.Given A(circle) find the A(triangle formed by the center and the endpoints of the sector).In $triangle ABC$, I is the incenter. Area of $triangle IBC = 28$, area of $triangle ICA= 30$ and area of $triangle IAB = 26$. Find $AC^2 − AB^2$Given the length of two altitudes and one side , find the area of triangle.problem about Circumscribed circle of triangleGeometry: Finding the length of a segment formed in a circle-tangent problemFind the area of the circle inscribed in the smaller part of the sectorThe Problem of Triangle in GeometryFind the minimal length of a right triangle with altitude 1
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A problem related to circle , altitude , triangle.
The 2019 Stack Overflow Developer Survey Results Are InFinding value of an angle in a triangle.In △ABC, median AM = 17, altitude AD = 15 and the circum-radius R = 10. Find BC^2A circle is inscribed in sector of another bigger circle.Given A(circle) find the A(triangle formed by the center and the endpoints of the sector).In $triangle ABC$, I is the incenter. Area of $triangle IBC = 28$, area of $triangle ICA= 30$ and area of $triangle IAB = 26$. Find $AC^2 − AB^2$Given the length of two altitudes and one side , find the area of triangle.problem about Circumscribed circle of triangleGeometry: Finding the length of a segment formed in a circle-tangent problemFind the area of the circle inscribed in the smaller part of the sectorThe Problem of Triangle in GeometryFind the minimal length of a right triangle with altitude 1
$begingroup$
Consider a $triangle ABC.$ Draw circle $S$ such that it touches side $AB$ at $A$. This circle
passes through point $C$ and intersects segment $BC$ at $E.$
If Altitude $AD =frac21(sqrt3−1)sqrt2;$
and $;angle EAB = 15^circ,$ find AC.
Here is a sketch that I made:
Here , we can calculate AC by Pythagorean theorem if we find DC . This is where I'm stuck . How can I utilize the information that angle EAB is 15 deg ?
(This is not class-homework , I'm solving sample questions for a competitive exam )
geometry triangles
edited Mar 30 at 20:51
Glorfindel
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
$endgroup$
add a comment |
$begingroup$
Consider a $triangle ABC.$ Draw circle $S$ such that it touches side $AB$ at $A$. This circle
passes through point $C$ and intersects segment $BC$ at $E.$
If Altitude $AD =frac21(sqrt3−1)sqrt2;$
and $;angle EAB = 15^circ,$ find AC.
Here is a sketch that I made:
Here , we can calculate AC by Pythagorean theorem if we find DC . This is where I'm stuck . How can I utilize the information that angle EAB is 15 deg ?
(This is not class-homework , I'm solving sample questions for a competitive exam )
geometry triangles
edited Mar 30 at 20:51
Glorfindel
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
$endgroup$
$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03
add a comment |
$begingroup$
Consider a $triangle ABC.$ Draw circle $S$ such that it touches side $AB$ at $A$. This circle
passes through point $C$ and intersects segment $BC$ at $E.$
If Altitude $AD =frac21(sqrt3−1)sqrt2;$
and $;angle EAB = 15^circ,$ find AC.
Here is a sketch that I made:
Here , we can calculate AC by Pythagorean theorem if we find DC . This is where I'm stuck . How can I utilize the information that angle EAB is 15 deg ?
(This is not class-homework , I'm solving sample questions for a competitive exam )
geometry triangles
edited Mar 30 at 20:51
Glorfindel
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
$endgroup$
Consider a $triangle ABC.$ Draw circle $S$ such that it touches side $AB$ at $A$. This circle
passes through point $C$ and intersects segment $BC$ at $E.$
If Altitude $AD =frac21(sqrt3−1)sqrt2;$
and $;angle EAB = 15^circ,$ find AC.
Here is a sketch that I made:
Here , we can calculate AC by Pythagorean theorem if we find DC . This is where I'm stuck . How can I utilize the information that angle EAB is 15 deg ?
(This is not class-homework , I'm solving sample questions for a competitive exam )
geometry triangles
geometry triangles
edited Mar 30 at 20:51
Glorfindel
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
edited Mar 30 at 20:51
Glorfindel
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
edited Mar 30 at 20:51
Glorfindel
3,41381930
edited Mar 30 at 20:51
Glorfindel
3,41381930
edited Mar 30 at 20:51
Glorfindel
3,41381930
3,41381930
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
asked Mar 27 '14 at 16:53
A GooglerA Googler
1,72032143
1,72032143
$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03
add a comment |
$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03
$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03
$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.
So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $measuredangle AFG=15°$ and $measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=ACcdotsin15°$ which gives you
$$AC=fracADsin 15°=42$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
$endgroup$
add a comment |
$begingroup$
Below is a scanned image of my solution worked out on paper.
answered Apr 3 '14 at 2:05
AjayAjay
1126
$endgroup$
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
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
$begingroup$
As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.
So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $measuredangle AFG=15°$ and $measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=ACcdotsin15°$ which gives you
$$AC=fracADsin 15°=42$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
$endgroup$
add a comment |
$begingroup$
As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.
So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $measuredangle AFG=15°$ and $measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=ACcdotsin15°$ which gives you
$$AC=fracADsin 15°=42$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
$endgroup$
add a comment |
$begingroup$
As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.
So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $measuredangle AFG=15°$ and $measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=ACcdotsin15°$ which gives you
$$AC=fracADsin 15°=42$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
$endgroup$
As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.
So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $measuredangle AFG=15°$ and $measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=ACcdotsin15°$ which gives you
$$AC=fracADsin 15°=42$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
answered Mar 27 '14 at 19:02
MvGMvG
31.2k450107
31.2k450107
add a comment |
add a comment |
$begingroup$
Below is a scanned image of my solution worked out on paper.
answered Apr 3 '14 at 2:05
AjayAjay
1126
$endgroup$
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
add a comment |
$begingroup$
Below is a scanned image of my solution worked out on paper.
answered Apr 3 '14 at 2:05
AjayAjay
1126
$endgroup$
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
add a comment |
$begingroup$
Below is a scanned image of my solution worked out on paper.
answered Apr 3 '14 at 2:05
AjayAjay
1126
$endgroup$
Below is a scanned image of my solution worked out on paper.
answered Apr 3 '14 at 2:05
AjayAjay
1126
edited Apr 3 '14 at 3:07
answered Apr 3 '14 at 2:05
AjayAjay
1126
answered Apr 3 '14 at 2:05
AjayAjay
1126
answered Apr 3 '14 at 2:05
AjayAjay
1126
1126
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
add a comment |
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
1
1
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
It would be helpful if instead of using periods, you could meet the character limit by giving a brief description of the image (even just "below is a scanned image of my solution worked out on paper"). This makes certain elements of the site (such as flags and review queues) which only use text work better, and makes the site more accessible to people with impaired vision.
$endgroup$
– Alex Becker
Apr 3 '14 at 2:53
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
OK I will do so
$endgroup$
– Ajay
Apr 3 '14 at 3:07
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
$begingroup$
done. Thanks for suggestion.(again character limit)
$endgroup$
– Ajay
Apr 3 '14 at 3:08
add a comment |
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$begingroup$
Isn't the description of $A$ and $C$ asymmetric? You don't think there's a difference between touching and passing through?
$endgroup$
– user2345215
Mar 27 '14 at 17:03