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Thread: [ASK] P to AC

  1. #1
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    [ASK] P to AC

    Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ....
    A. $\displaystyle 2\sqrt3$ cm
    B. $\displaystyle 3\sqrt3$ cm
    C. $\displaystyle 2\sqrt6$ cm
    D. $\displaystyle 3\sqrt6$ cm
    E. $\displaystyle 4\sqrt6$ cm

    So AH is $\displaystyle 8\sqrt2$ cm and PH = $\displaystyle 6\sqrt2$. What do I do now? Is the triangle HPC a right triangle?
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  2. #2
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    Re: [ASK] P to AC

    I cannot visualize this. You are going to have to give us some relation between the vertices. Give the cube an orientation and tell us which corner corresponds to which letter.
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  3. #3
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    Re: [ASK] P to AC

    ABCD is the base, while EFGH is the top. AB is the side closer to the monitor, ordered clockwise.
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  4. #4
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    Re: [ASK] P to AC

    Quote Originally Posted by Monoxdifly View Post
    ABCD is the base, while EFGH is the top. AB is the side closer to the monitor, ordered clockwise.
    Ok, so E is above A, F is above B, G is above C, and H is above D. Then this problem has no solution.
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  5. #5
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    Re: [ASK] P to AC

    Huh? Why?
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    Re: [ASK] P to AC

    Quote Originally Posted by Monoxdifly View Post
    Huh? Why?
    Because that is how geometry works. Why would you think this problem does have a solution? You are not giving us images. I asked where the vertices were in relation to each other, and you gave me four vertices and then said the rest are above it. So, I assigned them however I wanted. The assignment I gave has no solution. Unless you want to describe where the upper vertices are, this problem has no solution.
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  7. #7
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    Re: [ASK] P to AC

    Do you know a site where we can make a cube and label them online?
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  8. #8
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    Re: [ASK] P to AC

    There are 8 vertices. Here were the assignments I made:

    Front bottom right: A
    Front bottom left: B
    Back bottom left: C
    Back bottom right: D
    Front top right: E
    Front top left: F
    Back top left: G
    Back top right: H

    Just tell me your labeling. Be specific.
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  9. #9
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    Re: [ASK] P to AC

    Front bottom left: A
    Front bottom right: B
    Back bottom right: C
    Back bottom left: D
    Front top left: E
    Front top right: F
    Back top right: G
    Back top left: H
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  10. #10
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    Re: [ASK] P to AC

    Ok, that is actually a very similar layout to what I suggested. Here is the problem. If P is "within" AB, I assume that means it is a point on the line segment between A and B (along the edge of the cube). This means that AP is less than AB. AH has lengths $8\sqrt{2}$. Triangle APH is a right triangle (with PAH being the right angle). This means that PH has length $\sqrt{128+|\text{AP}|^2}$ So, you have the formula:

    $$|\text{AP}| = 3\sqrt{128+|\text{AP}|^2}$$

    Squaring both sides, you get:

    $$|\text{AP}|^2 = 9(128+|\text{AP}|^2)$$

    $$|\text{AP}|^2 = -72$$

    This is not possible. The square of a real number is always nonnegative. So either this problem has no solution over the reals, or you did not copy it correctly.
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  11. #11
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    Re: [ASK] P to AC

    Sorry, I did a typo. It should be "If P is within AH".
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  12. #12
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    Re: [ASK] P to AC

    Quote Originally Posted by Monoxdifly View Post
    Sorry, I did a typo. It should be "If P is within AH".
    Consider triangle ACH. Each side has length $8\sqrt{2}$. Therefore, it is an equilateral triangle. Since $|\text{AP}| = 3|\text{PH}|$ and $|\text{AP}|+|\text{PH}| = 4|\text{PH}| = 8\sqrt{2}$, we have $|\text{PH}| = 2\sqrt{2}$ and $|\text{CH}| = 8\sqrt{2}$. Then by the Law of Cosines, we have:

    $$|\text{PC}|^2 = 128+8-128\cos 60^\circ = 136-64 = 72$$

    $$|\text{PC}| = \sqrt{72} = 6\sqrt{2}$$

    Can you finish from here? The final answer is $3\sqrt{6}$.
    Last edited by SlipEternal; Apr 8th 2018 at 07:06 PM.
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    Re: [ASK] P to AC

    Where did you get 128 before the cos? Shouldn't $\displaystyle 2\times2\sqrt2\times8\sqrt2$ be 64?
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  14. #14
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    Re: [ASK] P to AC

    Quote Originally Posted by Monoxdifly View Post
    Where did you get 128 before the cos? Shouldn't $\displaystyle 2\times2\sqrt2\times8\sqrt2$ be 64?
    You are correct. I realized after that I didn't need that length to determine the answer. If $|\text{AP}|=6\sqrt{2} $ and PAH is a 60 degree angle, then P to AC is $6\sqrt{2}\dfrac{\sqrt{3}}{2}=3\sqrt{6} $
    Thanks from Monoxdifly
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  15. #15
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    Re: [ASK] P to AC

    Oh, right. Sine function. How could I forget trigonometry at a time like this. Thanks for the help.
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