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This problem states that on $△ABC$, $AB=3$, $AC=1$, $AD=1$ $∠BAC=θ$. It says that there are two points that fullfill the condition $DP_0=\frac{1}{3}BC$ and asks for the range of $\cos \theta$ that is needed to fullfill the condition. These images are given:

enter image description here

Image 1, Image 2

enter image description here The problem says that by looking at image#2, it can be known that the conditions for P to exist are $0\leq AP<AP_0$ and the explanation of the problem says that this is because $P$ is between $A$ and $P_0$ but I don't understand this. How can I get the conditions for $P$ to exist only by looking at the second image?

ラゴズ
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  • Please take time to define $P_0$ as the point on $AC$ with the property $AP_0:AC=AD:AB=1:3$, then introduce the set of all points $P$, so that $P$ is on... (?? which line / segment ??) and on the circle centered in $D$ with radius $DP_0$ (i assume). One point with this property seems to be $P_0$. Now we need conditions for $\theta$ so that an other point $P\ne 0$ with the given properties should exist. Please make clear which are these properties, and always use only objects already introduced, it is very confusing to use "two points", tacitly both denoted by $P$, then write for both $P_0$. – dan_fulea Mar 03 '21 at 15:10

1 Answers1

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Applying law of cosine in $\triangle ABC$,

$BC^2 = 3^2 + 1^2 - 2 \cdot 3 \cos \theta = 10 - 6 \cos \theta$

Now in $\triangle ADP$, if $AP = x$,

$DP^2 = \frac{BC^2}{9} = 1^2 + x^2 - 2 x \cos \theta$

i.e $ \ \displaystyle \frac{10 - 6 \cos \theta}{9} = 1 - \cos^2 \theta + (x- \cos \theta)^2$

$(1 - 3 \cos \theta)^2 = 9 (x-\cos\theta)^2$

$1 - 3 \cos \theta = \pm (3x - 3 \cos \theta)$

So, $x = \frac{1}{3} $ or $x = \frac{6\cos\theta - 1}{3}$

Range of $\theta$ so we have two points on line segment $AC$ meeting the condition $DP = \frac{BC}{3}$,

we must have $0 \lt x \lt 1 \implies \frac{1}{6} \lt \cos\theta \lt \frac{2}{3}$.

Also note that when $\cos \theta = \frac{1}{3}$, we have only one point $P$ where $DP = \frac{BC}{3}$.

Math Lover
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