Question

If $\Delta ABC$ is inscribed in semicircle whose diameter is $AB$, then $\overline{AC}+\overline{BC}$ must be:

• A. equal to $\overline{AB}$
• B. equal to $\overline{AB}\sqrt{2}$
• C. $\ge AB\sqrt{2}$
• D. $\le \overline{AB}\sqrt{2}$

Answer 1

The correct option is D $\le \overline{AB}\sqrt{2}$

Because AB is the diameter of the semi-circle,

it follows that $\angle C=90$.

Now we can try to eliminate all the solutions except for one by giving counterexamples.

$\text{(A):}$ Set point C anywhere on the perimeter of the semicircle except on AB.

By triangle inequality, $AC+BC>AB$, so $\text{(A)}$ is wrong.

$\text{(B):}$ Set point C on the perimeter of the semicircle infinitesimally close to AB,

and so AC+BC almost equals AB, therefore $\text{(B)}$ is wrong.

$\text{(C):}$ Because we proved that AC+BC can be very close to AB in case $\text{(B)}$,

it follows that $\text{(C)}$ is wrong.

$\text{(E):}$ Because we proved that AC+BC can be very close to AB

in case $\text{(B)}$, it follows that $\text{(E)}$ is wrong.

Therefore, the only possible case is $\text{(D)}\le AB\sqrt{2}$