Difference between revisions of "2018 AIME II Problems/Problem 12"
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Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>. | Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>. | ||
− | ==Solution== | + | ==Solution 1== |
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes | For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes | ||
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This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <math>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</math> and hence <math>\left[ABCD\right] = 56 + 56 = \boxed{112}</math>. -kgator | This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <math>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</math> and hence <math>\left[ABCD\right] = 56 + 56 = \boxed{112}</math>. -kgator | ||
+ | ==Solution 2 (Another way to get the middle point)== | ||
+ | So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}</cmath>. Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2</cmath>. So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2</cmath>. So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2</cmath>. As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}</cmath>. Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}</cmath>. Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we can find out that <cmath>S_{3}=S_{4}</cmath>. So <math>P</math> is the middle point of <math>\overline {AC}</math> | ||
{{AIME box|year=2018|n=II|num-b=11|num-a=13}} | {{AIME box|year=2018|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:00, 6 October 2018
Problem
Let be a convex quadrilateral with
,
, and
. Assume that the diagonals of
intersect at point
, and that the sum of the areas of triangles
and
equals the sum of the areas of triangles
and
. Find the area of quadrilateral
.
Solution 1
For reference, , so
is the longest of the four sides of
. Let
be the length of the altitude from
to
, and let
be the length of the altitude from
to
. Then, the triangle area equation becomes
.
What an important finding! Note that the opposite sides and
have equal length, and note that diagonal
bisects diagonal
. This is very similar to what happens if
were a parallelogram with
, so let's extend
to point
, such that
is a parallelogram. In other words,
and
. Now, let's examine
. Since
, the triangle is isosceles, and
. Note that in parallelogram
,
and
are congruent, so
and thus
. Define
, so
. We use the Law of Cosines on
and
:
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot
on
gives
and therefore
. Seeing that
, we conclude that
is a 3-4-5 right triangle, so
. Then, the area of
is
. Since
, points
and
are equidistant from
, so
and hence
. -kgator
Solution 2 (Another way to get the middle point)
So, let the area of triangles
,
,
,
. Suppose
and
, then it is easy to show that
. Also, because
, we will have
. So
. So
. So
. So
. As a result,
. Then, we have
. Combine the condition
, we can find out that
. So
is the middle point of
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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