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Geometry problem - areas of triangles (contest math)

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Geometry problem - areas of triangles (contest math)

1

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Vasya

asked 4 hours ago

VasyaVasya

4,1351618

$endgroup$

add a comment | 

1

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Vasya

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

share|cite|improve this question

edited 3 hours ago

Vasya

asked 4 hours ago

VasyaVasya

4,1351618

$endgroup$

share|cite|improve this question

edited 3 hours ago

Vasya

asked 4 hours ago

VasyaVasya

4,1351618

share|cite|improve this question

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

1 Answer
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oldest

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4

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

user10354138user10354138

7,4722925

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

user10354138user10354138

7,4722925

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 4 hours ago

user10354138user10354138

7,4722925

$endgroup$

share|cite|improve this answer

answered 4 hours ago

user10354138user10354138

7,4722925

answered 4 hours ago

user10354138user10354138

7,4722925

answered 4 hours ago

user10354138user10354138

7,4722925