We have explored squares, circles, and triangles, but now it is time to tackle something a little more rugged! With a series of zig-zagging peaks and valleys is a mountain whose base length is still unknown- at least to us until we figure it out. Just like our previous puzzles, underneath the puzzle lies beautiful geometry.

We have a mountain with a base angle of $45^\circ$ . The surface of the mountain then zigzags through four distinct segments with lengths of $6, 5, 4,\$ and $3$ . Every single peak and valley forms a perfect $90^\circ$ right angle. Finally, the rightmost edge of the mountain drops straight down with a vertical height of $10$ . The problem asks us to find the base length.

Let’s look at two different ways to solve this—direct calculation with a step-by-step route, and then a nice solution by one of our followers!

1. Solution 1: By step by step calculation

Because every angle in the zig-zag is $90^\circ$ , and the starting angle is $45^\circ$ , every single segment of this mountain is moving at exactly a $45^\circ$ angle relative to the ground! This means we can break the mountain down into a series of $45^\circ - 45^\circ - 90^\circ$ triangles.

In a $45^\circ-45^\circ-90^\circ$ triangle, if the hypotenuse is $L$ , both the horizontal width and vertical height are exactly $\frac{L}{\sqrt 2}$ . This is a key point that will allow us to figure out the base length.

Look at the marked figure below.

The total base length is made up of the smaller segments $AB, \, BC, \, CD, \, DE, \, EF,$ and $FG$ .

Consider the leftmost triangle. It is a $45^\circ - 45^\circ - 90^\circ$ triangle with height $10$ . So, $AB=10$ .
To find $CD$ , look at the triangle with hypotenuse $6$ . As we stated above, it has a base length of $\frac{6}{\sqrt 2}$ . Therefore, $CD = \frac{6}{\sqrt 2}$ .
Similarly, we can find that;

$DE = \frac{5}{\sqrt 2}$

$EF = \frac{4}{\sqrt 2}$

$FG = \frac{3}{\sqrt 2}$

Now, it only remains to figure out the length of the segment $BC$ . Notice that $BC=h$ , which is the height of the smallest top triangular portion immediately after the large left triangle. If we figure out $h$ , then it suffices us to proceed.

Consider the triangle with hypotenuse $6$ . The smallest triangle is embedded in this bigger triangle with a hypotenuse of $6$ . For convenience, look at the zoomed-in figure below.

The right vertical leg $MN$ has length $\frac{3}{\sqrt 2}$ . The middle vertical leg $OP$ has length $\frac{4}{\sqrt 2}$ and $OQ$ has length $\frac{5}{\sqrt 2}$ . So, the segment $PQ$ has a length of $\frac{1}{\sqrt 2}$ .

Now consider the leftmost vertical leg $RU$ . It has a length of $\frac{6}{\sqrt 2}$ . And it is formed by the three segments $RS, \, ST,$ and $TU$ .

Notice that $RS$ is our unknown height $h$ , $ST$ has a height of $\frac{3}{\sqrt 2}$ (being in line with the segment $MN$ ), and $TU$ has a height of $\frac{1}{\sqrt 2}$ (being in line with the segment $PQ$ ).

Now, if we equate the above relations, we get the following :

\frac{6}{\sqrt 2} = h + \frac{3}{\sqrt 2}+\frac{1}{\sqrt 2} = h+\frac{4}{\sqrt 2}

Therefore,

h = \frac{6}{\sqrt 2} - \frac{4}{\sqrt 2} = \frac{2}{\sqrt 2}.

And finally, we can now get the total base length by adding up all the lengths of the horizontal segments.

\text{Total base}  =AB+BC+CD+DE+EF+FG

\text{Total base} = 10+h+\frac{6}{\sqrt 2}+\frac{5}{\sqrt 2}+\frac{4}{\sqrt 2}+\frac{3}{\sqrt 2}

Substitute the value for $h$ and simplify:

\text{Total base} = 10+\frac{2}{\sqrt 2}+\frac{6}{\sqrt 2}+\frac{5}{\sqrt 2}+\frac{4}{\sqrt 2}+\frac{3}{\sqrt 2}.

Then,

\text{Total base} = 10+\frac{20}{\sqrt 2} = 10+10\sqrt 2.

Thus, the mountain has a base length of $10+10\sqrt 2$ units.

2. Solution 2: An elegant approach

The next solution we have is a fascinating solution given by one of our follower’s (Hữu Hiếu) comment on our Facebook page. Read further to enjoy its beauty.

Let the two ends of the base be $A$ and $B$ , as shown above. Also, label the peaks $X, \, Y,$ and $Z$ . Now, let’s do some construction.

Extend the sides $AX$ and $BZ$ such that they meet at the point $P$ . Now, $ABP$ forms a triangle with $45^\circ - 45^\circ - 90^\circ$ angles. Furthermore, construct rectangles $XQYR$ and $RSZT$ as shown in the diagram given below. Then, we can use the relationship between opposite sides of a rectangle to proceed further.

Since $XQ = 6$ , we know its opposite side, $YR = 6$ as well.
Similarly, the total segment $RS = RY + YS$ . Since the mountain segment $YS = 4$ , we have $RS = 6 + 4 = 10$ . Therefore, its opposite side $TZ = 10$ .

Now, observe that the top triangle $TZP$ is also a $45^\circ - 45^\circ - 90^\circ$ right-angled triangle with hypotenuse $ZP$ and $TZ = 10$ . Recall that in a $45^\circ - 45^\circ - 90^\circ$ angled triangle, the two legs adjacent to the right angle are equal in length. Therefore, $TP = 10$ as well!

Now apply the Pythagoras' Theorem:

ZP^2=10^2+10^2 = 200.

So,

ZP = \pm \sqrt {200}.

Since we consider the positive value of ZP, we finally get that,

ZP = \sqrt{200} = 10\sqrt 2.

Finally, to get the base length, consider our larger triangle $ABP$ . Since $ABP$ is also a $45^\circ - 45^\circ - 90^\circ$ triangle, its height is exactly equal to its base length.

The height of the triangle is

BP = BZ+ZP = 10+10\sqrt 2.

Hence, the base length is also $10+10\sqrt 2$ units!

Related Articles

The Zig-Zag Mountain

1. Solution 1: By step by step calculation

2. Solution 2: An elegant approach