AIME 2019 I · 第 3 题

Solution 1

We know the area of the hexagon $ABCDEF$ to be $\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED$. Since $PR^2+RQ^2=PQ^2$, we know that $\triangle PRQ$ is a right triangle. Thus the area of $\triangle PQR$ is $150$. Since $\triangle PQR$ is a right triangle, we also know that $\sin \angle PQR = \frac{15}{25} = \frac{3}{5}$. Then the area of $\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}$. Preceding in a similar fashion for $\triangle PAF$, the area of $\triangle PAF$ is $10$. Since $\angle ERD = 90^{\circ}$, the area of $\triangle RED=\frac{25}{2}$. Thus our desired answer is $150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}$

Solution 2

Let $R$ be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$, and $F=(0,10)$. Using the shoelace theorem, the area is $\boxed{120}$. Shoelace theorem:Suppose the polygon $P$ has vertices $(a_1, b_1)$, $(a_2, b_2)$, ... , $(a_n, b_n)$, listed in clockwise order. Then the area of $P$ is

$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$ You can also go counterclockwise order, as long as you find the absolute value of the answer.

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Solution 3 (Easiest, uses only basic geometry too)

Note that $\triangle{PQR}$ has area $150$ and is a $3$-$4$-$5$ right triangle. Then, by similar triangles, the altitude from $B$ to $QC$ has length $3$ and the altitude from $A$ to $FP$ has length $4$, so $[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30$, meaning that $[ABCDEF]=\boxed{120}$. -Stormersyle

Solution 4

Knowing that $\triangle{PQR}$ has area $150$ and is a $3$-$4$-$5$ triangle, we can find the area of the smaller triangles $\triangle{DRE}$, $\triangle{APF}$, and $\triangle{CQB}$ and subtract them from $\triangle{PQR}$ to obtain our answer. First off, we know $\triangle{DRE}$ has area $12.5$ since it is a right triangle. To the find the areas of $\triangle{APF}$ and $\triangle{CQB}$ , we can use Law of Cosines ($c^2 = a^2 + b^2 - 2ab\cos C$) to find the lengths of $AF$ and $CB$, respectively. Computing gives $AF = \sqrt{20}$ and $CB = \sqrt{10}$. Now, using Heron's Formula, we find $\triangle{APF} = 10$ and $\triangle{CQB} = 7.5$. Adding these and subtracting from $\triangle{PQR}$, we get $150 - (10 + 7.5 + 12.5) = \boxed{120}$ -Starsher

Solution 5 (Official MAA)

Triangle $PQR$ is a right triangle with are $\tfrac12\cdot15\cdot20=150$. Each of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ shares an angle with $\triangle PQR$. Because the area of a triangle with sides $a,\,b,$ and included angle $\gamma$ is $\tfrac12a\cdot b\cdot \sin\gamma,$ it follows that the areas of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ are each $5\cdot5\cdot\tfrac{150}{ab},$ where $a$ and $b$ are the lengths of the sides of $\triangle PQR$ adjacent to the shared angle. Thus the sum of the areas of $\triangle PAF,\triangle QCB,$ and $\triangle RED$ is

$$5\cdot5\cdot\frac{150}{15\cdot25}+5\cdot5\cdot\frac{150}{25\cdot20}+5\cdot5\cdot\frac{150}{20\cdot15}=25\left(\frac25+\frac3{10}+\frac12\right)=30.$$ Therefore $ABCDEF$ has area $150-30=\boxed{120}$.

Solution 6 (Using simple trigonometry)

Let's say that $\angle APF$ = $u$. Then, we know that $\sin(u) = 20/25 = 4/5$. Therefore, the area of $\triangle{APF}$ is $0.5*5*5*\sin(u)=10$. Now, let's do the same thing with $\triangle{BQC}$. If we name $\angle BQC$ as $w$, we know that $\sin(w) = 15/25 = 3/5$. Therefore, the area of $\triangle{BQC}$ is $0.5*5*5*\sin(w)=7.5$. Since $\triangle{ERD}$ is a right-angled isosceles triangle, the area of $\triangle{ERD} = 12.5$. In conclusion, the area of $ABCDEF$ is $\triangle{PRQ}-(\triangle{APF}+\triangle{BQC}+\triangle{ERD})=150-(10+7.5+12.5)=\boxed{120}$.

Video Solution #1(Complementary Area Counting?)

https://youtu.be/JQdad7APQG8?t=417

Video Solution

https://www.youtube.com/watch?v=4jOfXNiQ6WM

Video Solution 2

https://youtu.be/TSKcjht8Rfk?t=941

~IceMatrix

Video Solution 3

https://youtu.be/9X18wCiYw9M

~Shreyas S