Tetrahedron $A B C D$ has base $\triangle A B C$. Point $E$ is the midpoint of $A B$. Point $F$ is on $A D$ so that $F D=2 A F$, point $G$ is on $B D$ so that $G D=2 B G$, and point $H$ is on $C D$ so that $H D=2 C H$. Point $M$ is the midpoint of $F G$ and point $P$ is the point of intersection of the line segments $E H$ and $C M$. What is the ratio of the volume of tetrahedron $E B C P$ to the volume of tetrahedron $A B C D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_b6707dc71c8b7ccd30afg-5.jpg?height=802&width=1136&top_left_y=431&top_left_x=554)
The solution will use two facts. The first is a general fact about triangles.

Fact 1: Suppose $\triangle Q R S$ has $X$ on $Q R$ and $Y$ on $S R$ so that $X R=2 Q X$ and $Y R=2 S Y$. If $Z$ is the point of intersection of $Q Y$ and $S X$ and $W$ is the point where the extension of $R Z$ intersects $Q S$, then $W R=5 W Z$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-18.jpg?height=396&width=544&top_left_y=1255&top_left_x=842)

The second fact that we will use is specific to the question.

Fact 2: The point $M$ is on the line segment connecting $D$ to $E$ so that $M D=2 E M$. (Note: this justifies the fact that $E H$ and $C M$ intersect as implied by the problem statement.)

We will answer the question before proving the two facts.

To start, we will show that the altitude of tetrahedron $A B C D$ is 5 times that of $E B C P$.

Let $K$ be the point where the line through $D P$ intersects $\triangle A B C$ and let $L$ be the point in $\triangle A B C$ so that $D L$ is perpendicular to $\triangle A B C$, and let $V$ be the point on $\triangle A B C$ so that $P V$ is perpendicular to $\triangle A B C$.

By construction, $D L$ is the altitude of $A B C D$ from $A$ to $\triangle A B C$ and $P V$ is the altitude of $E B C P$ from $P$ to $\triangle E B C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-19.jpg?height=498&width=290&top_left_y=171&top_left_x=972)

By Fact 2, $M$ is on $D E$ so that $M D=2 E M$ and $H$ is on $D C$ so that $H D=2 C H$ by assumption.

By Fact $1, K D=5 K P$.

Since $\triangle K P V$ and $\triangle K D L$ share an angle at $K$ and both have a right angle, the triangles are similar.

This means $\frac{P V}{D L}=\frac{K P}{K D}=\frac{1}{5}$.

Thus, the length of the altitude of $A B C D$ is five times that of $E B C P$.

Since $E$ is the midpoint of $A B$, the area of $\triangle E B C$ is half that of $\triangle A B C$.

Thus, the base area of tetrahedron $E B C P$ is half that of tetrahedron $A B C D$, while the heights are in ratio $1: 5$, which means that the volumes are in the ratio $1: 10$.

Now for the proofs of the facts.

Proof of Fact 1

We will first show that the height of $\triangle Q Z S$ from $Z$ is one fifth the height of $\triangle Q R S$ from $R$.

Suppose $\triangle Q Z S$ has height $h_{1}$ from $Z$ and $\triangle Q R S$ has height $h_{2}$ from $R$. The first goal is to show that $\frac{h_{1}}{h_{2}}=\frac{1}{5}$.

The area of $\triangle Q Z S$ is $\frac{1}{2}(Q S) h_{1}$ and the area of $\triangle Q R S$ is $\frac{1}{2}(Q S) h_{2}$.

Thus, if we take the ratio of their areas we get

$$
\frac{\frac{1}{2}(Q S) h_{1}}{\frac{1}{2}(Q S) h_{2}}=\frac{h_{1}}{h_{2}}
$$

and so the ratio of their heights is the same as the ratio of their areas.

This means we can show that $\frac{h_{1}}{h_{2}}=\frac{1}{5}$ by showing that the area of $\triangle Q R S$ is 5 times the area of $\triangle Q Z S$.

To do this, draw a line from $R$ to $Z$ and label the areas of $\triangle Q X Z, \triangle R X Z, \triangle R Y Z, \triangle S Y Z$, and $\triangle Q Z S$ by $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-19.jpg?height=385&width=526&top_left_y=2228&top_left_x=859)

We will now make several observations about the areas $A_{1}$ through $A_{5}$.

Since $\triangle Q X Z$ and $\triangle R X Z$ have the same altitude from $Z$ to $Q R$ and their bases satisfy $X R=2 Q X$, we get that

$$
A_{2}=2 A_{1}
$$

By similar reasoning, we also get the equations

$$
\begin{aligned}
A_{3} & =2 A_{4} \\
A_{2}+A_{3}+A_{4} & =2\left(A_{1}+A_{5}\right)=2 A_{1}+2 A_{5} \\
A_{1}+A_{2}+A_{3} & =2\left(A_{4}+A_{5}\right)=2 A_{4}+2 A_{5}
\end{aligned}
$$

Substituting Equation (1) into Equation (3) gives $A_{2}+A_{3}+A_{4}=A_{2}+2 A_{5}$ which simplifies to $A_{3}+A_{4}=2 A_{5}$.

Substituting Equation (2) into Equation (4) gives $A_{1}+A_{2}+A_{3}=A_{3}+2 A_{5}$ which simplifies to $A_{1}+A_{2}=2 A_{5}$.

Adding $A_{3}+A_{4}=2 A_{5}$ and $A_{1}+A_{2}=2 A_{5}$ gives $A_{1}+A_{2}+A_{3}+A_{4}=4 A_{5}$.

Adding $A_{5}$ to both sides gives $A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=5 A_{5}$, which exactly says that the area of $\triangle Q R S$ is five times that of $\triangle Q Z S$.

We will now let $T$ be the point on $Q S$ so that $R T$ is perpendicular to $Q S$ and let $U$ be on $W T$ so that $Z U$ is perpendicular to $R W$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-20.jpg?height=496&width=290&top_left_y=1259&top_left_x=972)

Note that $Z U$ is the height of $\triangle Q Z S$ from $Z$ and $R T$ is the height of $\triangle Q R S$ from $R$, so we know that $R T=5 Z U$.

Since $\angle W U Z=\angle W T R=90^{\circ}$ and $\angle Z W U=\angle R W T, \triangle Z W U$ is similar to $\triangle R W T$.

Since $\frac{Z U}{R T}=\frac{1}{5}$, we get that $\frac{W Z}{W R}=\frac{1}{5}$, which proves the fact.

## Proof of Fact 2

Consider $\triangle A D B$ and $\triangle F D G$. It is given that $\frac{F D}{A D}=\frac{2}{3}=\frac{G D}{B D}$, and since the triangles have a common angle at $D$, they are similar by side-angle-side similarity.

Let $N$ be the point at which $D E$ intersects $F G$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-21.jpg?height=399&width=699&top_left_y=169&top_left_x=775)

Since $\triangle A D B$ is similar to $\triangle F D G, \angle D F N=\angle D A E$.

We also have that $\angle F D N=\angle A D E$ since they are the same angle, which implies that $\triangle F D N$ is similar to $\triangle A D E$ since if two triangles have two angles in common, they must have three angles in common.

Since $\frac{F D}{A D}=\frac{2}{3}$, similarity implies $\frac{F N}{A E}=\frac{2}{3}$.

By similar reasoning, $\frac{G N}{B E}=\frac{2}{3}$.

Rearranging these two equations leads to $F N=\frac{2}{3} A E$ and $G N=\frac{2}{3} B E$, but $E$ is the midpoint of $A B$, so $A E=B E$, and hence, $F N=G N$.

This means $N$ is the midpoint of $F G$, so $M=N$, which shows that $M$ is on $D E$ since $N$ is on $D E$.

Finally, since $\triangle F D M$ and $\triangle A D E$ are similar, that $\frac{M D}{E D}=\frac{F D}{A D}=\frac{2}{3}$, which means that $M$ is on $D E$ so that $M D=2 E M$.

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)