Track / Stage 4 / 241 of 340 #501 of 2000

Problem 501

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

$17 \cdot 118$ In an isosceles $\triangle ABC$ , $AD$ is the altitude on the base $BC$ , and $BE$ is the altitude on $AC$ . Also, $AD$ and $BE$ intersect at $H$ . Meanwhile, $EF \perp BC$ intersects $BC$ at $F$ , and $M$ is a point on the extension of $AD$ such that $DM = EF$ . Also, $N$ is the midpoint of $AH$ . Let $MN^2 = b$ , $BN^2 = m$ , and $BM^2 = n$ . The relationship between $b$ , $m$ , and $n$ is
(A) $bm + n$ .
(D) The size of $b$ and $m + n$ is uncertain.
(Anhui Province, China Junior High School Mathematics Competition, 1996)

Your answer

Official solution

[Solution] Without loss of generality, take the triangle as an isosceles right triangle, i.e., $\triangle ABC$ (as shown in the figure). At this time, points $A, E, H, N$ coincide, and points $D, F$ coincide.

It is obvious that $\triangle BMA$ is the same as $\triangle NBM$ , both being isosceles triangles, i.e., $\angle NBM = 90^{\circ}$ .
Thus, we know that $\quad b = m + n$ .
Therefore, the answer is $(B)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.