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# geometry

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ABCD is a square.  Find the area of triangle AMP.

Apr 16, 2022

#1
+2540
+1

Extend point P so it is perpendicular to AB. Label the intersection point F.

$$[AMP] = {2 \over3} \times PF \div 2$$

$$\triangle APF$$ is similar to $$\triangle ABN$$. Because of this, $$PF = {1 \over3}$$

Thus, $$[AMP] = {2 \over 3} \times {1 \over 3} \div 2 = \color{brown}\boxed{1 \over 9}$$

Apr 16, 2022
#3
+2540
0

Wait... I made an error, I wrongfully concluded that PF was 1/3 BN. MaxWong's answer is correct, and mine is incorrect

BuilderBoi  Apr 16, 2022
#2
+9461
+2

Here I provide a solution with coordinate geometry.

Let D = (0, 0), C = (2, 0), A = (0, 2), B = (2, 2), M = (2/3, 2), N = (2, 1).

The equation of DM is:

$$y = \dfrac{2 - 0}{\dfrac23 - 0}x\\$$

i.e., $$y = 3x$$

The equation of AN is:

$$y - 1 = \dfrac{2 - 1}{0 - 2}(x - 2)\\ -2(y - 1) = x - 2\\ x + 2y-4=0$$

We solve for the coordinates of P.

$$\begin{cases}y = 3x\\x+2y-4=0\end{cases}\\ x + 2(3x) - 4 = 0\\ x= \dfrac47\\ y = 3\left(\dfrac47\right) = \dfrac{12}7$$

Hence P = (4/7, 12/7).

Therefore,

$$\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12\cdot AM \cdot \left(2 - \dfrac{12}7\right) \\ &=& \dfrac12 \cdot \dfrac23 \cdot \dfrac27\\ &=& \dfrac2{21} \end{array}$$

Apr 16, 2022
#4
+9461
+1

Here's an alternative solution, that may be easier to understand.

$$\begin{array}{rcl} \tan \angle BAN &=& \dfrac{1}{2}\\ \angle BAN &=& \arctan \dfrac12 \end{array}$$

$$\begin{array}{rcl} \tan \angle AMD &=& \dfrac{2}{\frac23}\\ &=& 3\\ \angle AMD &=& \arctan 3 \end{array}$$

We now use the sine rule:

$$\begin{array}{rcl} \dfrac{MP}{\sin \angle BAN} &=& \dfrac{AM}{\sin\left(180^\circ - \arctan \dfrac12 - \arctan 3\right)}\\ \sqrt5 MP&=& \dfrac{5 \sqrt 2}7\cdot\dfrac23\\ MP &=& \dfrac{2\sqrt{10}}{21} \end{array}$$

Then,

$$\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12 \cdot \dfrac23 \cdot MP \cdot \sin \angle AMP\\ &=& \dfrac{2\sqrt{10}}{63}\cdot \dfrac{3\sqrt{10}}{10}\\ &=& \dfrac2{21} \end{array}$$

Apr 16, 2022