Geometry problem - areas of triangles (contest math) The Next CEO of Stack OverflowContest...
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Geometry problem - areas of triangles (contest math)
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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 3 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 $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.
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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active
oldest
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active
oldest
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active
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votes
$begingroup$
Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.
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 $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.
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 $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.
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 $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.
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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