In triangle $ABC$, let $I$ be the incenter of triangle $ABC$. The line through $I$ parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB = 5$, $AC = 5$, and $BC = 8$, then find the area of triangle $AMN$.

ABJeIIy Apr 18, 2024

#1**0 **

Since I is the incenter, then ∠BIM=∠CIN=90∘. Also, since MN is parallel to BC, then ∠AMN=∠ABC and ∠ANM=∠ACB.

By the Law of Cosines on triangle ABC, [\cos C = \frac{AB^2 + AC^2 - BC^2}{2AB \cdot AC} = \frac{5^2 + 5^2 - 8^2}{2 \cdot 5 \cdot 5} = -\frac{3}{25}.]Then sinC=1−cos2C=2546.

pen_spark

Let x=BM and y=CN. Then by the Law of Sines on triangle ABM, [\frac{BM}{\sin C} = \frac{AB}{\sin \angle BAM} = \frac{AB}{\sin 90^\circ} = AB,]so BM=5⋅sinC=546.

Similarly, by the Law of Sines on triangle ACN, [\frac{CN}{\sin C} = \frac{AC}{\sin \angle CAN} = \frac{AC}{\sin 90^\circ} = AC,]so CN=5⋅sinC=546.

Then triangle AMN is a right triangle with legs of length 546, so its area is [\frac{1}{2} \cdot \frac{4 \sqrt{6}}{5} \cdot \frac{4 \sqrt{6}}{5} = \boxed{\frac{32}{5}}.]

bader Apr 18, 2024