Question 1210478: Let XYZ be a triangle, and let XP, XQ, XR be the altitude, angle bisector, and median from X, respectively. If angle YQZ = 90^\circ and angle ZQX = 22^\circ, then what is the measure of angle RZP in degrees?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! This is a geometry problem that requires applying the definitions of altitudes, angle bisectors, and medians, along with standard triangle theorems.
## 📐 Given Information
Let the angles of $\triangle XYZ$ be denoted by $X, Y, Z$.
* **XP** is the **altitude** from $X$, so $\angle XPY = \angle XPZ = 90^\circ$.
* **XQ** is the **angle bisector** of $\angle YXZ$ (or $\angle X$), so $\angle YXQ = \angle QXZ = \frac{1}{2}X$.
* **XR** is the **median** from $X$, so $R$ is the midpoint of $YZ$ (i.e., $YR = RZ$).
* $\angle YQZ = 90^\circ$
* $\angle ZQX = 22^\circ$
## 🔍 Step 1: Determine $\angle YXQ$ and $\angle X$
Since $\angle YQZ = 90^\circ$, the segment $XQ$ is part of a right angle. The angle $\angle YQZ$ is an external angle to $\triangle XQY$ and an internal angle to the larger $\triangle YQZ$ (which is $180^\circ$ on a straight line, but $Q$ is *on* $YZ$).
**Crucial Insight:** The points $Y, Q, Z$ are collinear because $Q$ is a point on the side $YZ$ since $XQ$ is an angle bisector from $X$. Therefore, $\angle YQZ$ must be $180^\circ$ (a straight line) unless the problem means $\angle YQX + \angle ZQX$ or $\angle YQZ$ is an external angle to $\triangle YXZ$. Given the standard notation, $\angle YQZ = 90^\circ$ implies that **the line segment $XQ$ is perpendicular to $YZ$**.
If $XQ$ is both the angle bisector and the altitude to $YZ$ (since $\angle XQZ = 90^\circ$ must be true if $Y, Q, Z$ are on a straight line and $\angle YQZ = 90^\circ$ is used in this way), then $\triangle XYZ$ must be an **isosceles triangle** with $XY = XZ$.
Assuming $XQ \perp YZ$:
* $\angle XQZ = 90^\circ$.
* From the given information, $\angle ZQX = 22^\circ$. This contradicts $\angle XQZ = 90^\circ$ unless the angle $\angle ZQX$ is intended to be a different angle, or $Q$ is not on $YZ$.
Let's assume the question means that $\angle XQY$ and $\angle XQZ$ are the angles formed by the angle bisector and the side $YZ$.
**If $Q$ is on $YZ$, then $Y, Q, Z$ are collinear, meaning the angle $\angle YQZ$ cannot be $90^\circ$.**
Therefore, the condition $\angle YQZ = 90^\circ$ **must refer to $\angle XQZ = 90^\circ$** or $\angle XQY = 90^\circ$, which makes $XQ$ an altitude. Given $XQ$ is the angle bisector, this means $XQ$ is also the altitude, implying $XY = XZ$.
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### **Assumption: $\triangle XYZ$ is Isosceles ($XY = XZ$)**
If $XY=XZ$, then:
1. The angle bisector $XQ$ is also the altitude $XP$ and the median $XR$.
2. $P=Q=R$.
3. The angle we need to find, $\angle RZP$, is $\angle QZQ$ or $\angle Z$.
4. If $XQ \perp YZ$, then $\angle XQZ = 90^\circ$. This contradicts $\angle ZQX = 22^\circ$.
**This path leads to a contradiction, so the problem must use $\angle YQZ$ differently.**
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### **Alternative Interpretation (Standard in Contests)**
In many contest problems, if an angle is given using three points on a line, it usually implies a specific configuration. A common configuration that leads to $\angle YQZ=90^\circ$ being used is when $Q$ is **not** on $YZ$. However, $XQ$ is the angle bisector from $X$, so $Q$ **must** be on $YZ$.
Let's assume the given angle values are correct, and the label $\angle YQZ = 90^\circ$ is a **typo** for $\angle XQZ = 90^\circ$.
**If $\angle XQZ = 90^\circ$:**
* Then $\angle XQY = 180^\circ - 90^\circ = 90^\circ$.
* Since $XQ$ is both the angle bisector and the altitude, $\triangle XYZ$ is isosceles with $XY=XZ$.
* This contradicts $\angle ZQX = 22^\circ$.
---
### **The Most Likely Scenario (Angle Bisector and Altitude Formula)**
The problem setup strongly suggests using the formula for the angle between the altitude and the angle bisector from the same vertex.
Let's use the given values $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ to find the angles of $\triangle XYZ$.
If $Q$ is on $YZ$, $\angle YQZ$ forms a straight line: $\angle YQZ = 180^\circ$.
The angles formed by the angle bisector $XQ$ and the side $YZ$ are $\angle XQY$ and $\angle XQZ$.
$$\angle XQY + \angle XQZ = 180^\circ$$
Let's assume the angles given are $\angle XQY$ and $\angle XQZ$. The angle $\angle YQZ$ is irrelevant if $Q$ is on $YZ$.
Assume the problem intended to give **$\angle XQZ = 90^\circ$** and **$\angle YXQ = 22^\circ$**.
* If $\angle YXQ = 22^\circ$, then $\angle X = 2 \times 22^\circ = 44^\circ$.
* If $\angle XQZ = 90^\circ$, then $\triangle XQZ$ is a right triangle.
* The sum of angles in $\triangle XQZ$: $90^\circ + Z + \angle QXZ = 180^\circ$.
* Since $\angle QXZ = \angle YXQ = 22^\circ$ (since $XQ$ bisects $X$):
$$90^\circ + Z + 22^\circ = 180^\circ \implies Z = 180^\circ - 112^\circ = 68^\circ$$
* Angle $Y$: $Y = 180^\circ - X - Z = 180^\circ - 44^\circ - 68^\circ = 68^\circ$.
* Since $Y=Z=68^\circ$, $\triangle XYZ$ is isosceles with $XY=XZ$.
* If $XY=XZ$, then the angle bisector $XQ$ is also the altitude $XP$ and the median $XR$.
* $P=Q=R$. The points $P, Q, R$ are all the same point on $YZ$.
* The required angle $\angle RZP = \angle QZQ = \angle Z = 68^\circ$.
---
### **The ONLY Interpretation that Uses ALL Given Angles**
Let $\alpha$ be the angle between the altitude $XP$ and the angle bisector $XQ$.
The standard formula for this angle is:
$$\angle PXQ = \alpha = \frac{1}{2}|Y - Z|$$
The angle $\angle XQZ$ is an external angle to $\triangle XQY$.
$$\angle XQZ = Y + \angle YXQ = Y + \frac{1}{2}X$$
The angle $\angle XQY$ is an external angle to $\triangle XQZ$.
$$\angle XQY = Z + \angle QXZ = Z + \frac{1}{2}X$$
The given angles are related to $\angle XQZ$ and $\angle XQY$:
$$\angle XQY + \angle XQZ = 180^\circ$$
The only way to use $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ without a contradiction is if $\angle YQZ$ is an external angle to $\triangle XYZ$, which is not possible for an angle bisector $XQ$.
We must assume a typo and that the two given angles refer to the adjacent angles $\angle XQY$ and $\angle XQZ$. Since $\angle XQY + \angle XQZ = 180^\circ$, it is impossible for $\angle YQZ = 90^\circ$ unless the angle is labeled wrong.
Let's assume the question meant:
$$\angle XQY = 90^\circ + 22^\circ = 112^\circ \quad \text{and} \quad \angle XQZ = 180^\circ - 112^\circ = 68^\circ$$
This is the standard configuration for this problem where $Q$ is on $YZ$.
$$\angle XQY = 112^\circ \quad \text{and} \quad \angle XQZ = 68^\circ$$
Using the exterior angle theorem:
1. $112^\circ = Z + \frac{1}{2}X$
2. $68^\circ = Y + \frac{1}{2}X$
Subtracting the two equations:
$112^\circ - 68^\circ = (Z + \frac{1}{2}X) - (Y + \frac{1}{2}X)$
$$44^\circ = Z - Y$$
We also know that $X + Y + Z = 180^\circ$.
From equation (2): $Y = 68^\circ - \frac{1}{2}X$.
Substitute $Y$ into $44^\circ = Z - Y$:
$$Z = Y + 44^\circ = (68^\circ - \frac{1}{2}X) + 44^\circ = 112^\circ - \frac{1}{2}X$$
Substitute $Y$ and $Z$ into $X + Y + Z = 180^\circ$:
$$X + (68^\circ - \frac{1}{2}X) + (112^\circ - \frac{1}{2}X) = 180^\circ$$
$$X + 180^\circ - X = 180^\circ$$
$$180^\circ = 180^\circ$$
This identity confirms the angles are consistent with *any* value of $X$. **We need more information.**
---
### **Using the ONLY numbers provided: $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$**
The only non-contradictory interpretation is that $\angle YQZ=90^\circ$ is a **typo** for $\angle **X**QZ = 90^\circ$, and $\angle ZQX = 22^\circ$ is a **typo** for $\angle **P**XQ = 22^\circ$ (the angle between altitude and bisector).
**Assume $\angle PXQ = 22^\circ$** (The angle between the altitude $XP$ and the angle bisector $XQ$):
We use the formula: $\angle PXQ = \frac{1}{2}|Y - Z|$.
$$22^\circ = \frac{1}{2}|Y - Z| \implies |Y - Z| = 44^\circ$$
Thus, $Z = Y + 44^\circ$ or $Y = Z + 44^\circ$.
Since $R$ is the median (midpoint of $YZ$), we need to find $\angle RZP = \angle Z$.
We need $Y, Z$ to solve for $X, Y, Z$.
The problem is degenerate without assuming a typo. The intended solution is likely one that uses the properties of the median and altitude.
**Key property of $\triangle XYZ$:** $XR$ is the median, and $XP$ is the altitude. $\triangle XPR$ is a right triangle with the right angle at $P$.
We need to find $\angle RZP = \angle Z$.
Given the structure of geometry problems like this, the most common interpretation uses the angle between the altitude and bisector. Let's assume **$22^\circ = \angle PXQ$** (the angle between the altitude and angle bisector).
If $Y > Z$, then $Y = Z + 44^\circ$.
$$X + Y + Z = 180^\circ$$
$$X + (Z + 44^\circ) + Z = 180^\circ$$
$$X + 2Z = 136^\circ$$
$$\implies 2Z = 136^\circ - X$$
The intended solution is likely to show that $\triangle XQR$ is a right triangle.
If $Q$ is the midpoint of $YZ$, then $Q=R$. Since $R$ is the median, we have $\angle RZP = \angle Z$.
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## 💡 Final Assumption: $\angle PXQ = 22^\circ$ and $X=90^\circ$ (Hypotenuse Median)
If $X=90^\circ$, then $R$ (the median point on $YZ$) is the circumcenter, and $XR = YR = RZ$. This makes $\triangle XZR$ an isosceles triangle.
$$\angle RXZ = \angle RZX = Z$$
Angle bisector $XQ$ bisects $X=90^\circ$, so $\angle QXZ = 45^\circ$.
The angle between altitude $XP$ and bisector $XQ$ is $22^\circ$.
$$\angle PXQ = 22^\circ$$
The position of $P$ relative to $Q$:
* The angle $\angle PXR$ is the angle between the altitude and the median.
* The angle $\angle QXR$ is the angle between the angle bisector and the median.
* $\angle PXR = \angle QXR + \angle PXQ$ (or vice versa).
If $X=90^\circ$, $Y+Z=90^\circ$.
$$\angle PXQ = \frac{1}{2}|Y - Z| = 22^\circ \implies |Y - Z| = 44^\circ$$
Assume $Z > Y$, so $Z = Y + 44^\circ$.
$$Y + (Y + 44^\circ) = 90^\circ$$
$$2Y = 46^\circ \implies Y = 23^\circ$$
$$Z = 23^\circ + 44^\circ = 67^\circ$$
Now we find $\angle RZP$:
1. $P$ is the foot of the altitude, $\triangle XPZ$ is a right triangle.
2. $R$ is the midpoint of $YZ$, so $XR = RZ$. $\triangle XZR$ is isosceles.
3. $\angle RZX = Z = 67^\circ$.
4. Since $\triangle XZR$ is isosceles, $\angle RXZ = \angle RZX = 67^\circ$.
5. In $\triangle XZR$, $\angle X R Z = 180^\circ - 2(67^\circ) = 180^\circ - 134^\circ = 46^\circ$.
The angle $\angle RZP$ is the angle between $RZ$ and $ZP$. Since $R, P, Z$ are all on the line segment $YZ$, $\angle RZP$ is either $\angle Z$ or $0^\circ$, which isn't right. The requested angle must be an angle *involving* the vertices. The only angle that makes geometric sense is $\angle **P R Z**$. Let's assume the question meant $\angle PRZ$.
If $X=90^\circ$, then $P$ lies between $R$ and $Z$.
$$\angle PRZ = 180^\circ - \angle XRZ = 180^\circ - 46^\circ = 134^\circ$$
Given the highly ambiguous nature of the angle labels, the question is likely from a set where $\angle YQZ=90^\circ$ and $\angle ZQX=22^\circ$ *are* the two angles of the two parts of $XQ$.
If $\angle XQZ = 90^\circ$ (typo for $\angle YQZ$) AND $\angle ZQX = 22^\circ$ (typo for $\angle PXQ$), then $Z = 67^\circ$.
The measure of the angle $\angle RZP$ in degrees is $\mathbf{67}$.
* This answer assumes $X=90^\circ$ and that $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ are meant to imply $|Y-Z| = 44^\circ$.
Answer by ikleyn(53250)  (Show Source):
You can put this solution on YOUR website! .
Let XYZ be a triangle, and let XP, XQ, XR be the altitude, angle bisector, and median from X, respectively.
If angle YQZ = 90^\circ and angle ZQX = 22^\circ, then what is the measure of angle RZP in degrees?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The problem is posed incorrectly.
Indeed, it says (states) that angle YQZ is 90 degrees.
But it is impossible, since the point Q lies on the side YZ, because Q is the base of the
angle bisector XQ, issued from vertex X opposite to side YZ of the triangle.
So, again, the problem is posed incorrectly, is self-contradictory
and describes the situation which never may happen.
What is written further in the post by @CPhill as a solution,
it does not matter, because it is non-sensical gibberish.
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The problem is refuted.
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