This is a similar triangles problem. The walls are parallel (both vertical), so the lines of sight from the man's eye level create proportional triangles with the tops of the walls.

Assume the man's eye is 0.1 m below his top (negligible for most calculations, but included for precision). Let h be the man's height in meters.

The similar triangles give:

4 / (6 - h) = h / (6 - (h - 0.1))

Solving approximately (ignoring the 0.1 m for simplicity, as the man's height is much smaller):

4 / (6 - h) ≈ h / 6

4 * 6 ≈ h * (6 - h)

24 ≈ 6h - h²

h² - 6h + 24 = 0? Wait, cross-multiply correctly:

h * (6 - h) = 4 * 6 = 24

6h - h² = 24

h² - 6h + 24 = 0

Discriminant: 36 - 96 = -60 (impossible).

I see the issue— the man's head isn't at ground level. The proportion is between the full wall heights relative to the line distances.

The left wall's visible height is 6 m (full wall from ground), but since walls are parallel and man is between, it's the shadow or line to top.

It's a perspective problem with parallel lines, so the triangles from the eye to the tops are similar.

Let e = eye height ≈ h (since close).

Distance to first wall: 4 m, height above eye: 6 - e

Distance to second wall: 4 + 6 = 10 m from first? No, the bases are 6 m apart.

The diagram shows left wall base to man base: 4 m, right wall base to man base: 6 m, walls 6 m and 4 m tall? Wait, labels: left is 4 m tall, right 6 m tall.

Looking back at the image description:

• Left wall height: 4 m

• Right wall height: 6 m

• Distance from left wall to man: 4 m (horizontal)

• Distance from man to right wall: 6 m (horizontal)

Walls parallel and vertical, man upright.

Lines from man's eye to top of left wall and top of right wall.

To find h = man's height. Assume eye at top of head for simplicity, or approximately h.

The triangles are: small triangle to left wall, large to right.

But since parallel, the proportion of heights above eye level to distances.

Let e ≈ h (eye height).

The left triangle: base 4 m, height (4 - e)? No— the wall is 4 m tall, but the line is from eye to top of wall.

Actually, the full proportion: the line of sight crosses the ground at some point, but better: use intercept theorem or Thales' theorem for parallel lines.

Since walls are parallel, the ratios of distances from a point (man) to the lines.

The height of man is the "missing" part that makes the proportions equal.

Consider the full heights if man were not there, but the lines from eye create similar triangles with the walls as bases.

Left: distance 4 m, wall height 4 m

Right: distance 4+6=10 m from left, but from man 6 m, wall 6 m

The ratio for left: wall height / distance = 4 / 4 = 1

For right: 6 / 6 = 1

Same slope, so the lines are parallel! But in the diagram, they converge? No, if ratios equal, the lines of sight to tops would be parallel if same slope.

The walls are different heights but distances make ratio 1:1 for both.

To see the top of left wall at distance 4 m for 4 m height means slope = 4/4 = 1 (45 degrees).

For right wall, 6 m height at 6 m from man = slope 1.

So both lines of sight have the same slope from the man's eye.

Therefore, the man's eye must be positioned such that the slope to both tops is the same, meaning the eye is at the height where the 45-degree lines meet or something.

Since same slope, the eye height e above ground must make the height above eye to wall top over distance equal for both.

For left: wall height 4 m, distance 4 m left, so height above eye for left top: 4 - e, distance 4 m, tan(theta) = (4 - e)/4

For right: (6 - e)/6

Since the lines are drawn converging at eye, but for the view to see both tops, set equal:

(4 - e)/4 = (6 - e)/6

Cross-multiply: 6(4 - e) = 4(6 - e)

24 - 6e = 24 - 4e

-6e + 4e = 24 - 24

-2e = 0

e = 0

That can't be— if e=0, then slopes 4/4=1, 6/6=1, yes equal, but man height 0?

That doesn't make sense for the problem— the man has height ?, so probably the eye is not at h, but the lines are from the top of man or something.

Let's re-examine the diagram from the ASCII art and description.

The image shows:

• Left wall: labeled "4m" at base (height? or distance?)

The labels are:

• Under left wall: 4m (height of wall)

• Under man to left: 4m (distance)

• Under right wall: 6m (distance from man to right)

• Right wall height: 6m? The label "6m" is next to right wall upper.

From the text: "4m" next to left wall vertical, "6m" next to right wall vertical.

Yes, left wall 4 m tall, right wall 6 m tall.

Horizontal: left distance 4 m, right distance 6 m.

Man stands upright between them.

The lines are from the man's position to the top of each wall, crossing at the man's head/eye.

It's a standard "man's height in alley" problem using similar triangles, where the man's height corresponds to the ratio of distances.

The ground triangles.

The large triangle to the right wall: base from left wall to right wall = 4 + 6 = 10 m, height 6 m (right wall).

But the left wall is 4 m, so the man is blocking part.

The idea is that the line to the left top creates a triangle with base 4 m, height 4 m.

The line to the right top creates a large triangle with base 4 + 6 = 10 m, height = ? but to match the slope.

Since the slopes are different, the man's height is the difference where the lines would intersect above ground.

To see both tops, the eye is at the intersection point of the two lines from tops to ground or something.

Standard solution for such problems: the two walls are parallel, so the lines from their tops to the ground would intersect at infinity, but here the observer is between.

The problem is to find the man's height such that he can just see the tops with the lines tangent or from his eye.

But the diagram shows the man with ? at head, and lines from head to tops of both walls.

To "see" both, the eye is at the point where the line of sight is drawn, but to find h.

Use the property of similar triangles with the ground.

The triangle from the left wall top to the ground at the right.

But the standard way is to consider the proportion of the segments.

Since walls are parallel, the triangles formed by the line of sight and the ground are similar to the "air" triangles above the man.

Let's set it up mathematically.

Let the ground be x-axis, left wall at x=0, height 4 m at x=0.

Man at x=4 m, height h.

Right wall at x=10 m (4+6), height 6 m at x=10.

The line from top of left wall (0,4) to top of man's head (4, h).

The other line from top of right wall (10,6) to top of man's head (4,h).

The head is the intersection point of the two lines: line from (0,4) to (10,6), but no— the lines are from eye to tops, so the eye is below the tops, but the lines are the sight lines passing through the eye.

Since the man is upright, the eye is at (4, h) approximately, and the sight lines are from (4,h) to (0,4) and from (4,h) to (10,6).

That's always true for any h; you can always draw lines from any point to the tops.

So what is the puzzle? The ? is the height, but what's the condition?

Looking back at the image text: "the man stands upright."

"*both walls are vertical and parallel."

Then the diagram with lines from the man's head to the tops of both walls, crossing.

Perhaps the lines are drawn from the tops of the walls to the base, crossing at the man's head.

That makes sense for a puzzle: the lines from top of left wall to base of right wall, and from top of right wall to base of left wall, and they cross at the man's position, and the height at crossing is the man's height.

Yes, that's a classic geometry puzzle!

So, line 1: from top left (0,4) to base right (10,0)

Line 2: from