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I'm currently doing an exercise form my geometry book. The question is asking for the volume of the pyramid $N.ABCD$ (i.e. a pyramid of base $ABCD$ and with the tip $N$). The construction is as follows,

Let $ABCD.EFGH$ denote a cube. Let $M$ be the midpoint of the segment $BF$ and $N$ the intersection of $AM$ and $BE$. If the sides of the cube is $9$ cm. What is the volume of the pyramid $N.ABCD$?

Illustration of my construction

My goal here is trying to find $NN'$ (to be defined below).

I tried constructing a perpendicular from $N$ onto $AB$ and $BF$, lets call them $NN'$ and $NI$ respectively. If we let $MI$ be $y$ and $N'B$ be $x$. Through similar triangles $AMB$ and $NMI$ we obtain $\frac{y}{x}= \frac{1}{2}$.

I also found out that $NM = y \sqrt{5}$. As well as $AN = \sqrt{(9/2 -y)^{2} + (9-2y)^{2}}$. Now its easy to see that $AM = \frac{9 \sqrt{5}}{2}$, hence we obtain the gnarly equation,

$\sqrt{(9/2 -y)^{2} + (9-2y)^{2}} + y \sqrt{5} =\frac{9 \sqrt{5}}{2} $

Which is just ridiculous. I felt like something is missing here, can someone confirm this?

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  • $\begingroup$ The slope of $AM$ has half the magnitude of the slope of $BE$. What does that tell you about the relation between $x$ and $AB - x$? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ Is it $\frac{NN'}{AB-x} =2 \frac{NN'}{x}$ $\endgroup$ Commented 14 hours ago
  • $\begingroup$ You have the $2$ on the wrong side, and it's more complicatedly expressed than needed. It tells you $AB - x = 2x$, or $x = \frac{1}{3}AB$. You can find $NN'$ easily from that, can't you? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ I see, that'll give us $NN' = 3$. But how did you conclude $AM$ has a slope half of $BE$? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ I assume it's from the fact that $M$ is the median of $BF$ $\endgroup$ Commented 14 hours ago

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Your equation isn't gnarly nor ridiculous but it is redundant and circular.

$\sqrt{(9/2 -y)^{2} + (9-2y)^{2}} + y \sqrt{5} =\frac{9 \sqrt{5}}{2}$

$\sqrt{(\frac {9-2y}2)^2 + (9-2y)^2} + y\sqrt{5} =\frac {9\sqrt{5}}2$

$(9-2y)\sqrt{\frac 14 + 1} + y\sqrt{5} = \frac {9\sqrt 5}2$

$\frac {\sqrt 5}2 (9-2y) +y\sqrt{5} = \frac {9\sqrt 5}2$

$(9-2y) + 2y = 9$

$9=9$

That means you've overdone your solutions.

Instead note that that your similar triangles give you:

So $\frac {NN'}{AN'}=\frac {MB}{AB} = \frac 12$

And $\frac {NN'}{N'B} =\frac {AE}{AB} = 1$

And we have $AN' + N'B = AB=9$

So $NN' = \frac 12 AN'$

$NN' = N'B$

$AN' + N'B = 9$.

Just solve for $NN'$.

$AN' =2NN'$ and $N'B=NN'$ so $9 = AN' + N'B = 2NN' + NN' = 3NN'$ so $NN' = 3$.

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A purely synthetic approach:

Triangles $\triangle AEN$ and $\triangle MBN$ are similar. Since $AE=2BM,$ it follows that $EN=2BN$ and $BE=3BN.$

Then by similar right triangles, $N’N=\frac13AE=3.$

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When stuck in these kinds of problems, it can sometimes be useful to embed each point in a Cartesian plane. In our case, to compute $NN'$, we can embed $ABFE$ into a $2D$ Cartesian plane where the coordinates of each point is defined as follows : $A=(0, 0)$, $B=(L, 0)$, $F=(L, L)$ and $E=(0, L)$, where $L$ is the side length of the cube. It follows that $M=\left( L, \frac{L}{2} \right)$, and that the equations of the $(EB)$ and $(AM)$ lines are respectively $x \mapsto L - x$ and $x \mapsto \frac{x}{2}$. Therefore, the $x$-coordinate of the intersection point of $(EB)$ and $(AM)$ (i.e. $x_{N'}$) has to satisfy $L - x_{N'} = \frac{x_{N'}}{2}$, i.e. $x_{N'} = \frac{2}{3} L$, thus $NN' = y_{N'} = L - x_{N'} = \frac{x_{N'}}{2} = \frac{1}{3} L$.

And since the volume of a pyramid is given by $V = \frac{1}{3} \times Area(base) \times height$, the volume of the $N.ABCD$ pyramid is given by :

$$V_{N.ABCD}(L) = \frac{1}{3} \times Area(ABCD) \times NN' = \frac{1}{3} \times L^2 \times \left( \frac{1}{3} L \right) = \frac{1}{9} L^3$$

If $L = 9 \, \text{cm}$, then $NN' = 3 \, \text{cm}$, and the volume of the $N.ABCD$ pyramid is $81 \, \text{cm}^3$.

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In triangle $AFB$, point $N$ is the centroid of the triangle, because the intersection point $O$ of the diagonals of rectangle $ABFE$ is the midpoint of $AF$.

Therefore: $$\frac{AN}{AM} = \frac{2}{3}$$

From the similarity of right triangles $ANN'$ and $ABM$, we have: $$\frac{NN'}{MB} = \frac{AN}{AM} = \frac{2}{3}$$

Thus: $$NN' = \frac{2 \cdot MB}{3} = \frac{2 \cdot 4.5}{3} = 3 \text{ cm}$$

The volume of the pyramid is: $$V = \frac{1}{3} \cdot NN' \cdot S_{ABCD} = \frac{1}{3} \cdot 3 \cdot 9^2 = 81 \text{ cm}^3$$

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