MHB Solve $ac+bd-ad-bc$ Given $\dfrac{3}{4}$, $\sqrt{a^2+c^2}=15$

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Given the ratios of positive integers \( \frac{a}{c} = \frac{3}{4} \) and \( \frac{b}{d} = \frac{3}{4} \), the values of \( a \), \( b \), \( c \), and \( d \) can be expressed as \( a = 3k \), \( b = 3m \), \( c = 4k \), and \( d = 4m \) for some integers \( k \) and \( m \). The equation \( \sqrt{a^2+c^2} - \sqrt{b^2+d^2} = 15 \) can be simplified using these expressions, leading to \( \sqrt{25k^2} - \sqrt{25m^2} = 15 \). This results in \( 5k - 5m = 15 \), giving \( k - m = 3 \). Finally, substituting back into the expression \( ac + bd - ad - bc \) yields a specific value based on the integers derived from \( k \) and \( m \).
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Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and $$\sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15$$, find $ac+bd-ad-bc$.
 
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Re: Find ac+bd-ad-bc

anemone said:
Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and $$\sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15$$, find $ac+bd-ad-bc$.

Sqrt(a^2+c^2) = 5/4 c
Sqrt(b^2+d^2) = 5/4 d
so 5/4(c-d) = 15or c-d = 12
ac + bd – ad –bc
= ¾ c^2 + ¾ d^2 – ¾ cd – ¾ cd
= ¾(c-d)^2 = ¾ * 12^2 = 108
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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