Triangle Area and Slopes

Geometry Level 3

If x x is the length of one side of a triangle with a slope of zero, m 1 m_1 and m 2 m_2 are the slopes of the other two sides, and A A is the area of the triangle, then these four variables are related by the equation x P = Q ⋅ A ( m 1 R − m 2 S ) , x^P = Q \cdot A\big(m_1^R - m_2^S\big), where P P , Q Q , R R , and S S are non-zero integers.

Find ∣ P ∣ + ∣ Q ∣ + ∣ R ∣ + ∣ S ∣ . |P| + |Q| + |R| + |S|.


The answer is 6.

David Vreken by David Vreken , 42, USA

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1 solution

David Vreken
May 8, 2018

The area of △ X Y Z \triangle XYZ (with sides x x , y y , and z z , with x along the x-axis and Z Z at the origin) is A = 1 2 x y sin ⁡ Z A = \frac{1}{2}xy \sin Z .

By law of sines, y sin ⁡ Y = x sin ⁡ X \frac{y}{\sin Y} = \frac{x}{\sin X} , so y = x sin ⁡ Y sin ⁡ X y = \frac{x \sin Y}{\sin X} , which means A = x 2 sin ⁡ Y sin ⁡ Z 2 sin ⁡ X A = \frac{x^2 \sin Y \sin Z}{2 \sin X} .

By the angle sum of a triangle, X = 180 ° − ( Y + Z ) X = 180° - (Y + Z) , so sin ⁡ X = sin ⁡ ( 180 ° − ( Y + Z ) ) = sin ⁡ ( Y + Z ) \sin X = \sin (180° - (Y + Z)) = \sin (Y + Z) . Therefore, A = x 2 sin ⁡ Y sin ⁡ Z 2 sin ⁡ ( Y + Z ) A = \frac{x^2 \sin Y \sin Z}{2 \sin (Y + Z)} .

Using the sine addition identity, A = x 2 sin ⁡ Y sin ⁡ Z 2 ( sin ⁡ Y cos ⁡ Z + sin ⁡ Z cos ⁡ Y ) A = \frac{x^2 \sin Y \sin Z}{2 (\sin Y \cos Z + \sin Z \cos Y)} .

Rearranging, x 2 = 2 A ( cos ⁡ Z sin ⁡ Z + cos ⁡ Y sin ⁡ Y ) x^2 = 2A(\frac{\cos Z}{\sin Z} + \frac{\cos Y}{\sin Y}) or x 2 = 2 A ( 1 tan ⁡ Z + 1 tan ⁡ Y ) x^2 = 2A(\frac{1}{\tan Z} + \frac{1}{\tan Y}) .

Since m 1 = tan ⁡ Z m_1 = \tan Z and m 2 = tan ⁡ ( 180 ° − Y ) = − tan ⁡ Y m_2 = \tan (180° - Y) = -\tan Y , x 2 = 2 A ( 1 m 1 − 1 m 2 ) x^2 = 2A(\frac{1}{m_1} - \frac{1}{m_2}) or x 2 = 2 A ( m 1 − 1 − m 2 − 1 ) x^2 = 2A(m_1^{-1} - m_2^{-1}) .

This means P = 2 P = 2 , Q = 2 Q = 2 , R = − 1 R = -1 , and S = − 1 S = -1 , and ∣ P ∣ + ∣ Q ∣ + ∣ R ∣ + ∣ S ∣ = ∣ 2 ∣ + ∣ 2 ∣ + ∣ − 1 ∣ + ∣ − 1 ∣ = 6 |P| + |Q| + |R| + |S| = |2| + |2| + |-1| + |-1| = \boxed{6} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The tricky step was finally realizing that

m 1 + m 2 m 1 m 2 = m 1 − 1 + m 2 − 1 \dfrac{m_1 +m_2}{m_1 m_2} = m_1^{-1}+ m_2^{-1}

when once the area has been worked out first in this way

A = m 1 m 2 x 2 2 ( m 1 + m 2 ) A=\dfrac{m_1 m_2 x^2}{2(m_1 + m_2)}

Michael Mendrin - 3 years, 1 month ago

In general, it is not true that m 1 = tan ⁡ Z m_1 = \tan Z and m 2 = tan ⁡ Y m_2 = \tan Y .

Jon Haussmann - 3 years ago

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"Through trigonometry, the slope m m of a line is related to its angle of incline θ \theta by the tangent function m = tan ⁡ ( θ ) m= \tan(\theta) ." (See here .)

David Vreken - 3 years ago

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The problem is, in your solution, Y Y and Z Z are angles of the triangle (shown in red below).

The formula m = tan ⁡ θ m = \tan \theta only works when θ \theta is the angle the line makes with a horizontal line (shown in blue below). These are different angles.

Jon Haussmann - 3 years ago

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@Jon Haussmann – You are right! I edited my problem to restrict side x to have a horizontal slope.

David Vreken - 3 years ago

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