CEMC Fryer, 2016

Solution 1

In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$.

Since $BC=12$ and $D$ is the midpoint of $BC$, then $CD=\frac{12}{2}=6$.

In $\triangle ACD$, base $CD$ has length $6$, and corresponding height $AB$ has length $4$. (Since $\angle ABC=90^{\circ}$, $AB$ is the height of $\triangle ACD$ even though $AB$ is outside $\triangle ACD$.)

Thus, $\triangle ACD$ has area $\frac12(6)(4)=12$.

Solution 2

In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$.

Since $BC=12$ and $D$ is the midpoint of $BC$, then $CD=DB=6$.

In $\triangle ABD$, $AB=4$, $DB=6$, and $\angle ABD=90^{\circ}$, and so $\triangle ABD$ has area $\frac12(6)(4)=12$.

Similarly, $\triangle ABC$ has area $\frac12(12)(4)=24$, and so the area of $\triangle ACD$ is the area of $\triangle ABC$ minus the area of $\triangle ABD$, or $24-12=12$.

Solution 3

In $\triangle ABC$, $AB=4$, $BC=12$, and $\angle ABC=90^{\circ}$, and so $\triangle ABC$ has area $\frac12(12)(4)=24$.

A median of $\triangle ABC$ divides the triangle into two equal areas. Why?

In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$.

Therefore, $\triangle ACD$ and $\triangle ABD$ have equal bases ($CD=BD$).

Further, the height of $\triangle ABD$ is equal to the height of $\triangle ACD$ (both are $AB$).

Thus, $\triangle ACD$ and $\triangle ABD$ have equal bases and equal heights.

Since the area of each triangle equals one-half times the base times the height, then $\triangle ABD$ and $\triangle ACD$ have equal areas and so median $AD$ divides $\triangle ABC$ into equal areas.

Since $\triangle ABC$ has area 24, then $\triangle ACD$ has area $\frac{24}{2}=12$.
Solution 1

In $\triangle FSG$, $FS=18$, $SG=24$, and $\angle FSG=90^{\circ}$.

Thus, by the Pythagorean Theorem, $FG=\sqrt{18^2+24^2}=\sqrt{324+576}=\sqrt{900}=30$ (since $\text{FG gt;0}$).

Since $S$ is on $FH$ so that $FS=18$ and $SH=32$, then $FH=FS+SH=18+32=50$.

In $\triangle FGH$, $FH=50$, $FG=30$, and $\angle FGH=90^{\circ}$.

Thus, by the Pythagorean Theorem, $GH=\sqrt{50^2-30^2}=\sqrt{2500-900}=\sqrt{1600}=40$ (since $\text{GH gt;0}$).

In $\triangle FGH$, $FT$ is a median and so $T$ is the midpoint of $GH$.

In $\triangle FHT$, base $HT=\frac{40}{2}=20$, and height $FG=30$. (Since $\angle FGH=90^{\circ}$, $FG$ is the height of $\triangle FHT$ even though $FG$ is outside $\triangle FHT$.)

Thus, $\triangle FHT$ has area $\frac12(20)(30)=300$.

Solution 2

Since $S$ is on $FH$ so that $FS=18$ and $SH=32$, then $FH=FS+SH=18+32=50$.

In $\triangle FGH$, base $FH=50$, and height $SG=24$ (since $SG$ is perpendicular to $FH$, $SG$ is a height of $\triangle FGH$).

Thus, $\triangle FGH$ has area $\frac12(50)(24)=600$.

The median of a triangle divides the area of the triangle in half.

(Solution 3 to (a) shows an example of why a median divides a triangle’s area in half.)

Since $FT$ is a median of $\triangle FGH$, then the area of $\triangle FHT=\frac{600}{2}=300$.
We use the notation $|\triangle KLM|$ to represent the area of $\triangle KLM$, $|KPMQ|$ to represent the area of $KPMQ$, and so on.

In $\triangle KLM$, $KP$ is a median and so $2|\triangle KPM|=|\triangle KLM|$.

(Solution 3 to (a) shows an example of why a median divides a triangle’s area in half.)

In $\triangle KMN$, $KQ$ is a median and so $2|\triangle KMQ|=|\triangle KMN|$.

Therefore, $$|KLMN| = |\triangle KLM| + |\triangle KMN| = 2|\triangle KPM| + 2|\triangle KMQ|$$ and $$|KPMQ| = |\triangle KPM| + |\triangle KMQ|$$ which tells us that $|KLMN| = 2|KPMQ|$.

Since $|KPMQ| = 63$, then $|KLMN| = 2|KPMQ| = 2(63) = 126$.

Now $|KLMN| = |\triangle KRL| + |\triangle LRM| + |\triangle KRN| + |\triangle NRM|$.

Each of these four triangles is right-angled.

Since $KR = x$ and $LR = 6$, then $|\triangle KRL| = \frac{1}{2}x(6) = 3x$.

Since $LR = 6$ and $RM = 2x+2$, then $|\triangle LRM| = \frac{1}{2}(6)(2x+2) = 6x+6$.

Since $KR = x$ and $RN = 12$, then $|\triangle KRN| = \frac{1}{2}x(12) = 6x$.

Since $RN = 12$ and $RM = 2x+2$, then $|\triangle NRM| = \frac{1}{2}(12)(2x+2) = 12x+12$.

Therefore, $126 = 3x+(6x+6)+6x+(12x+12)$ or $126 = 27x + 18$ or $27x = 108$ and so $x = 4$.