1. ## [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?

2. ## 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.

3. ## Re: [ASK] P to AC

ABCD is the base, while EFGH is the top. AB is the side closer to the monitor, ordered clockwise.

4. ## Re: [ASK] P to AC

Originally Posted by Monoxdifly
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.

Huh? Why?

6. ## Re: [ASK] P to AC

Originally Posted by Monoxdifly
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.

7. ## Re: [ASK] P to AC

Do you know a site where we can make a cube and label them online?

8. ## 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.

9. ## 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

10. ## 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.

11. ## Re: [ASK] P to AC

Sorry, I did a typo. It should be "If P is within AH".

12. ## Re: [ASK] P to AC

Originally Posted by Monoxdifly
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 $3\sqrt{6}$.

13. ## Re: [ASK] P to AC

Where did you get 128 before the cos? Shouldn't $\displaystyle 2\times2\sqrt2\times8\sqrt2$ be 64?

14. ## Re: [ASK] P to AC

Originally Posted by Monoxdifly
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 $6\sqrt{2}\dfrac{\sqrt{3}}{2}=3\sqrt{6}$

15. ## 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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