A painting in an art gallery has height $h$ and is hung so that lower edge is a distance $d$ above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle $\theta$ subtended at his eye by the painting?)

A) $x = \sqrt { h ( d + h ) }$
B) $x = h ^ { 2 } ( h + d ) ^ { 2 }$
C) $x = d ^ { 2 } ( h + d ) ^ { 2 }$
D) $x = \sqrt { d ( h + d ) }$
E) $x = h ^ { 2 } ( h - d ) ^ { 2 }$

Question

A painting in an art gallery has height $h$ and is hung so that lower edge is a distance $d$ above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle $\theta$ subtended at his eye by the painting?)

A) $x = \sqrt { h ( d + h ) }$ 
B) $x = h ^ { 2 } ( h + d ) ^ { 2 }$ 
C) $x = d ^ { 2 } ( h + d ) ^ { 2 }$ 
D) $x = \sqrt { d ( h + d ) }$ 
E) $x = h ^ { 2 } ( h - d ) ^ { 2 }$

Quizplus · Accepted Answer

Let the observer stand at point O, and let the top of the painting be at point T. Then, we can draw a right triangle with vertices O, T, and the foot of the perpendicular from T to the wall, which we'll call F. Since angle OTF is a right angle, we have $$\tan \theta = \frac{TF}{OF}.$$ To maximize $\theta$, we want to maximize $\tan \theta$. From the right triangle OTF, we can see that $$\tan \theta = \frac{TF}{d + h}.$$ Thus, we want to maximize $TF$. Note that $TF$ is parallel to the ground and is the bottom side of another right triangle formed by extending the observer's line of sight until it hits the wall. Call the point where the observer's line of sight hits the wall point W. Then, we have another right triangle with vertices T, W, and the foot of the perpendicular from T to the ground, which we'll call G. Since $\angle WGT$ is a right angle, we have $$\tan \theta = \frac{TF}{WG} = \frac{TF}{x}.$$ Solving for $TF$ in terms of $\theta$ and substituting, we get $$\tan \theta = \frac{TF}{x} \Longrightarrow TF = x \tan \theta.$$ Thus, we want to choose $x$ to maximize $x \tan \theta$. By similar triangles (looking at right triangles OTF, OTG, and OTW), we have $$\frac{TF}{h} = \frac{x + d}{x} \Longrightarrow TF = h \cdot \frac{x + d}{x}.$$ Substituting into $x \tan \theta = TF$ and solving for $x$, we get $$x = \boxed{\sqrt{d(h+d)}}.$$

A Painting in an Art Gallery Has Height hhh And Is Hung So That Lower Edge Is a Distance

A Painting in an Art Gallery Has Height $h$ And Is Hung So That Lower Edge Is a Distance