Question 1210489: In triangle ABC, \angle A = 90^\circ. Altitude $\overline{AP},$ angle bisector $\overline{AQ},$ and median $\overline{AR}$ are drawn. If $PQ = 3$ and $QC = 4,$ find $AR.$
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! This is a challenging geometry problem that requires using properties of right triangles, angle bisectors, and medians.
## 1. Determine Side Lengths
In $\triangle ABC$, $\angle A = 90^\circ$.
### A. Locate Point C's Position
The angle bisector $\overline{AQ}$ divides $\overline{BC}$ into segments $\overline{BQ}$ and $\overline{QC}$.
We are given $QC = 4$. Let $BQ = y$.
The entire hypotenuse is $BC = BQ + QC = y + 4$.
### B. Use the Segment $PQ$
The altitude $\overline{AP}$ and the angle bisector $\overline{AQ}$ are distinct segments on the hypotenuse $\overline{BC}$.
We are given $PQ = 3$.
Since $P$ is the foot of the altitude from $A$, $\triangle APB$ and $\triangle APC$ are right triangles.
Since $Q$ is between $P$ and $C$ (altitude is often closer to the right angle vertex):
$$PC = PQ + QC = 3 + 4 = 7$$
Then, the segment $BP$ is $BC - PC$:
$$BP = (y + 4) - 7 = y - 3$$
### C. Apply the Geometric Mean Theorem (Leg Rule)
In right $\triangle ABC$:
1. $AC^2 = PC \cdot BC$ (Leg $\overline{AC}$ is the geometric mean of $\overline{PC}$ and $\overline{BC}$)
2. $AB^2 = BP \cdot BC$ (Leg $\overline{AB}$ is the geometric mean of $\overline{BP}$ and $\overline{BC}$)
Substitute the lengths:
1. $AC^2 = 7(y+4)$
2. $AB^2 = (y-3)(y+4)$
### D. Apply the Angle Bisector Theorem
The angle bisector $\overline{AQ}$ divides the opposite side $\overline{BC}$ into segments proportional to the other two sides:
$$\frac{AB}{AC} = \frac{BQ}{QC}$$
$$\frac{AB}{AC} = \frac{y}{4}$$
$$4 \cdot AB = y \cdot AC \quad \text{(Equation 1)}$$
### E. Solve for $y$ (the length of $BQ$)
Square Equation 1:
$$16 \cdot AB^2 = y^2 \cdot AC^2$$
Substitute the Leg Rule expressions for $AB^2$ and $AC^2$:
$$16 \cdot [(y-3)(y+4)] = y^2 \cdot [7(y+4)]$$
Since $y+4 = BC$ cannot be zero, we can divide both sides by $(y+4)$:
$$16(y-3) = 7y^2$$
$$16y - 48 = 7y^2$$
$$7y^2 - 16y + 48 = 0$$
Let's check the position of $P$ and $Q$. The altitude $\overline{AP}$ falls between the angle bisector $\overline{AQ}$ and the median $\overline{AR}$ if the triangle is not isosceles.
**Revisiting the Geometric Setup:** In a right triangle, the order of the points on the hypotenuse from $B$ to $C$ must be: **$P$ (Altitude), $Q$ (Angle Bisector), $R$ (Median)** *or* **$P, R, Q$** *or* **$R, Q, P$**, etc.
If $Q$ is between $P$ and $C$, our lengths are:
* $PC = 7$
* $BP = y-3$
* $BC = y+4$
The discriminant ($\Delta$) of the quadratic equation $7y^2 - 16y + 48 = 0$ is:
$$\Delta = (-16)^2 - 4(7)(48) = 256 - 1344 = -1088$$
Since the discriminant is negative, there are **no real solutions for $y$**. This means the geometric configuration $B, P, Q, C$ is impossible with the given lengths.
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## 2. Re-evaluate the Geometric Setup (P and Q Order)
Since $P$ and $Q$ must both be on the segment $\overline{BC}$, and the previous setup failed, let's assume the order of points is $B, Q, P, C$.
* $BC = y + 4$ (Still the total length)
* $BQ = y$
* $QC = 4$
* $PQ = 3$
If $Q$ is between $B$ and $P$:
$$BP = BQ + QP = y + 3$$
$$PC = BC - BP = (y+4) - (y+3) = 1$$
Now the geometric relationships are:
1. $AC^2 = PC \cdot BC = 1(y+4) = y+4$
2. $AB^2 = BP \cdot BC = (y+3)(y+4)$
Apply the Angle Bisector Theorem ($\frac{AB}{AC} = \frac{BQ}{QC}$):
$$\frac{AB^2}{AC^2} = \left(\frac{y}{4}\right)^2$$
$$16 \cdot AB^2 = y^2 \cdot AC^2$$
Substitute the Leg Rule expressions:
$$16 \cdot [(y+3)(y+4)] = y^2 \cdot [y+4]$$
Divide by $(y+4)$ (since $y+4 = BC \neq 0$):
$$16(y+3) = y^2$$
$$16y + 48 = y^2$$
$$y^2 - 16y - 48 = 0$$
### Solve the Quadratic Equation for $y$
Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$$y = \frac{16 \pm \sqrt{(-16)^2 - 4(1)(-48)}}{2(1)}$$
$$y = \frac{16 \pm \sqrt{256 + 192}}{2}$$
$$y = \frac{16 \pm \sqrt{448}}{2}$$
$$y = \frac{16 \pm \sqrt{64 \cdot 7}}{2} = \frac{16 \pm 8\sqrt{7}}{2}$$
Since $y$ must be positive (it is a length $BQ$):
$$y = 8 + 4\sqrt{7}$$
## 3. Find the Length of the Median $\overline{AR}$
The median $\overline{AR}$ goes from the vertex of the right angle to the midpoint ($R$) of the hypotenuse $\overline{BC}$.
### A. Find the length of the hypotenuse $BC$
$$BC = y + 4$$
$$BC = (8 + 4\sqrt{7}) + 4 = 12 + 4\sqrt{7}$$
### B. Apply the Median to Hypotenuse Theorem
In a right triangle, the length of the median to the hypotenuse is **half the length of the hypotenuse**.
$$AR = \frac{1}{2} BC$$
$$AR = \frac{1}{2} (12 + 4\sqrt{7})$$
$$AR = \mathbf{6 + 2\sqrt{7}}$$
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