AIME 2015 II · 第 7 题

Solution 1

If $\omega = 25$, the area of rectangle $PQRS$ is $0$, so

$$\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0$$ and $\alpha = 25\beta$. If $\omega = \frac{25}{2}$, we can reflect $APQ$ over $PQ$, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$,

$$[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90$$ so

$$45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}$$ and

$$\beta = \frac{180}{625} = \frac{36}{125}$$ so the answer is $m + n = 36 + 125 = \boxed{161}$.

Solution 2

Similar triangles can also solve the problem.

First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)

After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$. We then know that $CE = \frac{77}{5}$.

Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$.

Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$. Since $PF+FQ=PQ=\omega$. We can solve for $PF$ and $FQ$ in terms of $\omega$. We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$.

Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$. We can set up the proportion:

$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$. Solving for $AF$, $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$.

We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$.

Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$.

This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$.

Solution 3

Heron's Formula gives $[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,$ so the altitude from $A$ to $BC$ has length $\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.$

Now, draw a parallel to $AB$ from $Q$, intersecting $BC$ at $T$. Then $BT = w$ in parallelogram $QPBT$, and so $CT = 25 - w$. Clearly, $CQT$ and $CAB$ are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so

$$\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.$$ Solving gives $[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} \cdot w = \dfrac{36w}{5} - \dfrac{36w^2}{125}$, so the answer is $36 + 125 = 161$.

Solution 4

Using the diagram from Solution 2 above, label $AF$ to be $h$. Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$, so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$. Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$, you find $\frac{w}{25}=\frac{h}{36/5}$. Thus, $h=\frac{36w}{125}$. To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$, and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$.

Solution 5

Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$. Hence, $h=\frac{36}{5}$

Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$

Hence, $PS=\frac{3}{5}(12-L)$

Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$

Now, it is obvious that we want to find $L$ in terms of $W$.

Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$

Hence.. we can now set up similar triangles.

We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$.

Plugging back in..

$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$

Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$

Hence, $125+36=\boxed{161}$

Solution 6

Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$, so $\beta = \frac{36}{125}$, yielding $\boxed{161}$.

Solution 7(Quick and Easy)

Using the diagram in Solution 2, the area of $PQRS$ is ($w$)($PR$), we only need to find $PR$. Because $ABC$ and $APQ$ are similar in a ratio of $25:w.$ Because of $AB$:$AC$ $=12:15$ and $AP$:$AB$= $w:25,$ we can derive that $BP=\frac{(25-w)(12)}{25}.$ By using LoC on $ABC$, it is obvious that $cos(B)=\frac{4}{5}$ and $sin(B)=\frac{3}{5}.$ $PR=\frac{3bp}{5}=\frac{(25-w)(36)}{125}.$ Multiply $w$, we get $\frac{36W}{5}-\frac{36W^2}{125}$ which means the answer is $\boxed{161}$

fixed latex errors, ~eqb5000

Solution 7 (fake solve)

We are given that the area of rectangle $PQRS$ is a quadratic polynomial in terms of its width $\omega = PQ$. Let $h$ be the height of the rectangle (the length of $PS$ and $QR$). The area is given by $Area(\omega) = \omega h$.

Step 1: Identify the roots of the quadratic

A quadratic is uniquely determined by three points. We can identify two "degenerate" cases where the area of the rectangle is zero:

1. If $\omega = 0$, the rectangle collapses into a vertical line segment (the altitude), so $Area(0) = 0$.
2. If $\omega = 25$ (the length of base $BC$), the rectangle collapses into the segment $BC$ itself, meaning its height $h = 0$, so $Area(25) = 0$.

Since the roots are $0$ and $25$, the area polynomial must be of the form:

$Area(\omega) = k\omega(\omega - 25) = k\omega^2 - 25k\omega$

Comparing this to the given form $Area(\omega) = \alpha \omega - \beta \omega^2$, we see that $\beta = -k$.

Step 2: Find the altitude of triangle ABC

To find the constant $k$, we need one more non-zero point. First, we calculate the area of $\triangle ABC$ using Heron's Formula. The semi-perimeter is $s = \frac{12+25+17}{2} = 27$.

$Area(\triangle ABC) = \sqrt{27(27-12)(27-25)(27-17)} = \sqrt{27 \cdot 15 \cdot 2 \cdot 10} = 90$

Let $H$ be the altitude from $A$ to $BC$. Using $Area = \frac{1}{2}bh$:

$90 = \frac{1}{2}(25)(H) \implies H = \frac{180}{25} = \frac{36}{5}$

Step 3: Use similar triangles for a test point

Consider the case where $\omega = \frac{25}{2}$ (the width is half the base). By similar triangles ($\triangle APQ \sim \triangle ABC$), the height of the small triangle above the rectangle must be half of the total altitude $H$. Thus, the height of the rectangle $h$ is also half the total altitude:

$h = \frac{1}{2}H = \frac{18}{5}$

Our test point is $(\omega, Area) = \left(\frac{25}{2}, \frac{25}{2} \cdot \frac{18}{5}\right) = \left(\frac{25}{2}, 45\right)$.

Step 4: Solve for beta

Substitute the point into our quadratic equation:

$45 = k\left(\frac{25}{2}\right)\left(\frac{25}{2} - 25\right)$ $45 = k\left(\frac{25}{2}\right)\left(-\frac{25}{2}\right)$ $45 = -\frac{625}{4}k$ $k = -\frac{180}{625} = -\frac{36}{125}$

Since $\beta = -k$, we have $\beta = \frac{36}{125}$. The values $m=36$ and $n=125$ are relatively prime, so $m+n = 36 + 125 = \boxed{161}$.

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=554s

~MathProblemSolvingSkills.com