PRMO 2013 question 19
19. In a triangle \( ABC \) with \( \angle BCA = 90^\circ \), the perpendicular bisector of \( AB \) intersects segments \( AB \) and \( AC \) at \( X \) and \( Y \), respectively. If the ratio of the area of quadrilateral \( BXYC \) to the area of triangle \( ABC \) is \( 13 : 18 \) and \( BC = 12 \), then what is the length of \( AC \)?
Prerequisites:
1. If two triangles are similar, ratio of their areas is same as ratios of their sides squared.
Why?
Because their heights share the same ratio.
a1/a2 = h1/h2
So area ratio = a1h1/a2h2 = a1(a1h2/a2)/a2h2 = a1^2/a2^2
Solution:
Firstly, we can state that, \[ \frac{\text{Area of } A Y X}{\text{Area of } A B C} = \frac{5}{18} \tag{1} \] This is because it is given that, \[ \frac{\text{Area of } B X Y C}{\text{Area of } A B C} = \frac{13}{18} \] and, \[ \text{Area of } A B C = \text{Area of } A Y X + \text{Area of } B X Y C. \] **Next we can state that,** \[ \frac{A B^2}{4 \times A C^2} = \frac{5}{18} \tag{2} \] This is because \( A B C \) and \( A Y X \) are similar triangles. So the ratio of their areas is the same as the square of ratio of the lengths of their sides. This gives us \[ \frac{A X^2}{A C^2} = \frac{5}{18}. \] Add to this the fact that \( A X = \frac{A B}{2} \) is a given condition. **Next we can state that,** \[ \frac{A C^2 + 12^2}{4 \times A C^2} = \frac{5}{18} \tag{3} \] This is because, \[ A B^2 = A C^2 + 12^2 \] using Pythagoras theorem, and since we know that \( B C = 12 \). This can now be solved for \( A C \). **This gives us,** \[ A C = 36 \]
Another solution using Shoelace formula:
For a simple (non‑self‑intersecting) quadrilateral with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3), (x_4,y_4)\) listed in order around the perimeter (either clockwise or counter‑clockwise), the shoelace formula gives its area \(A\) as \[ A \;=\;\frac12\Bigl|\,(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)\;-\;(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\Bigr|. \] First, place \(C\) at the origin, \(B=(12,0)\), \(A=(0,a)\) so that \(\angle C=90^\circ\) and \(BC=12\), \(AC=a\). Then - The midpoint \(X\) of \(AB\) is \(\bigl(6,\tfrac a2\bigr)\). - The perpendicular bisector of \(AB\) has slope \(\tfrac{12}{a}\) and passes through \(X\), so its equation is \[ y-\frac a2=\frac{12}{a}\,(x-6). \] - Its intersection \(Y\) with \(AC\) (the \(y\)–axis, \(x=0\)) is \[ Y=\Bigl(0,\;\tfrac a2-\frac{72}{a}\Bigr). \] - The quadrilateral \(BXYC\) has vertices \((12,0),(6,\tfrac a2),(0,\tfrac a2-72/a),(0,0)\). By the shoelace formula its area is \[ \Area(BXYC) =\frac12\Bigl(9a-\tfrac{432}{a}\Bigr)\,, \] while \[ \Area(ABC)=\frac12\cdot12\cdot a=6a. \] So \[ \frac{\Area(BXYC)}{\Area(ABC)} =\frac{\frac12\bigl(9a-432/a\bigr)}{6a} =\frac34-\frac{36}{a^2} \stackrel!=\frac{13}{18}. \] Solving \[ \frac34-\frac{36}{a^2}=\frac{13}{18} \quad\Longrightarrow\quad \frac{36}{a^2}=\frac1{36} \quad\Longrightarrow\quad a^2=1296 \quad\Longrightarrow\quad a=36. \] Hence \(\displaystyle AC=36\).
Prerequisites:
1. If two triangles are similar, ratio of their areas is same as ratios of their sides squared.
Why?
Because their heights share the same ratio.
a1/a2 = h1/h2
So area ratio = a1h1/a2h2 = a1(a1h2/a2)/a2h2 = a1^2/a2^2
Solution:
Firstly, we can state that, \[ \frac{\text{Area of } A Y X}{\text{Area of } A B C} = \frac{5}{18} \tag{1} \] This is because it is given that, \[ \frac{\text{Area of } B X Y C}{\text{Area of } A B C} = \frac{13}{18} \] and, \[ \text{Area of } A B C = \text{Area of } A Y X + \text{Area of } B X Y C. \] **Next we can state that,** \[ \frac{A B^2}{4 \times A C^2} = \frac{5}{18} \tag{2} \] This is because \( A B C \) and \( A Y X \) are similar triangles. So the ratio of their areas is the same as the square of ratio of the lengths of their sides. This gives us \[ \frac{A X^2}{A C^2} = \frac{5}{18}. \] Add to this the fact that \( A X = \frac{A B}{2} \) is a given condition. **Next we can state that,** \[ \frac{A C^2 + 12^2}{4 \times A C^2} = \frac{5}{18} \tag{3} \] This is because, \[ A B^2 = A C^2 + 12^2 \] using Pythagoras theorem, and since we know that \( B C = 12 \). This can now be solved for \( A C \). **This gives us,** \[ A C = 36 \]
Another solution using Shoelace formula:
For a simple (non‑self‑intersecting) quadrilateral with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3), (x_4,y_4)\) listed in order around the perimeter (either clockwise or counter‑clockwise), the shoelace formula gives its area \(A\) as \[ A \;=\;\frac12\Bigl|\,(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)\;-\;(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\Bigr|. \] First, place \(C\) at the origin, \(B=(12,0)\), \(A=(0,a)\) so that \(\angle C=90^\circ\) and \(BC=12\), \(AC=a\). Then - The midpoint \(X\) of \(AB\) is \(\bigl(6,\tfrac a2\bigr)\). - The perpendicular bisector of \(AB\) has slope \(\tfrac{12}{a}\) and passes through \(X\), so its equation is \[ y-\frac a2=\frac{12}{a}\,(x-6). \] - Its intersection \(Y\) with \(AC\) (the \(y\)–axis, \(x=0\)) is \[ Y=\Bigl(0,\;\tfrac a2-\frac{72}{a}\Bigr). \] - The quadrilateral \(BXYC\) has vertices \((12,0),(6,\tfrac a2),(0,\tfrac a2-72/a),(0,0)\). By the shoelace formula its area is \[ \Area(BXYC) =\frac12\Bigl(9a-\tfrac{432}{a}\Bigr)\,, \] while \[ \Area(ABC)=\frac12\cdot12\cdot a=6a. \] So \[ \frac{\Area(BXYC)}{\Area(ABC)} =\frac{\frac12\bigl(9a-432/a\bigr)}{6a} =\frac34-\frac{36}{a^2} \stackrel!=\frac{13}{18}. \] Solving \[ \frac34-\frac{36}{a^2}=\frac{13}{18} \quad\Longrightarrow\quad \frac{36}{a^2}=\frac1{36} \quad\Longrightarrow\quad a^2=1296 \quad\Longrightarrow\quad a=36. \] Hence \(\displaystyle AC=36\).
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