Geometry problem - areas of triangles (contest math) The Next CEO of Stack OverflowContest Math GeometryMath contest geometry probabilitymath contest geometry proof problemMath contest geometry proof problem 2Contest Math Possible Triangles3D Geometry Contest Math Problemmath contest geometry problemInscribed and circumscribed non-regular polygonsSynthetic geometry with/without measurement vs analytic geometryRing Theoretical Method of Solving a Math Olympiad Problem
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Geometry problem - areas of triangles (contest math)
The Next CEO of Stack OverflowContest Math GeometryMath contest geometry probabilitymath contest geometry proof problemMath contest geometry proof problem 2Contest Math Possible Triangles3D Geometry Contest Math Problemmath contest geometry problemInscribed and circumscribed non-regular polygonsSynthetic geometry with/without measurement vs analytic geometryRing Theoretical Method of Solving a Math Olympiad Problem
$begingroup$
This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?
contest-math euclidean-geometry
asked 4 hours ago
VasyaVasya
4,1351618
$endgroup$
add a comment |
$begingroup$
This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?
contest-math euclidean-geometry
asked 4 hours ago
VasyaVasya
4,1351618
$endgroup$
add a comment |
$begingroup$
This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?
contest-math euclidean-geometry
asked 4 hours ago
VasyaVasya
4,1351618
$endgroup$
This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?
contest-math euclidean-geometry
contest-math euclidean-geometry
asked 4 hours ago
VasyaVasya
4,1351618
asked 4 hours ago
VasyaVasya
4,1351618
edited 4 hours ago
Vasya
asked 4 hours ago
VasyaVasya
4,1351618
asked 4 hours ago
VasyaVasya
4,1351618
asked 4 hours ago
VasyaVasya
4,1351618
4,1351618
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac12S$.
Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.
answered 4 hours ago
user10354138user10354138
7,4722925
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac12S$.
Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.
answered 4 hours ago
user10354138user10354138
7,4722925
$endgroup$
add a comment |
$begingroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac12S$.
Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.
answered 4 hours ago
user10354138user10354138
7,4722925
$endgroup$
add a comment |
$begingroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac12S$.
Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.
answered 4 hours ago
user10354138user10354138
7,4722925
$endgroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac12S$.
Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.
answered 4 hours ago
user10354138user10354138
7,4722925
answered 4 hours ago
user10354138user10354138
7,4722925
answered 4 hours ago
user10354138user10354138
7,4722925
answered 4 hours ago
user10354138user10354138
7,4722925
7,4722925
add a comment |
add a comment |
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